Memorator Algebră

clasele 5 - 8

≡ Memorator Algebră

∎ Introducerea factorilor sub radical

Exersează! - 2

A. Introduceți factorii sub radical:

a) \({7\sqrt{2}}\)

b) \({3\sqrt{3}}\)

c) \({6\sqrt{2}}\)

d) \({3^2\sqrt{2}}\)

e) \({2^2\sqrt{5}}\)

f) \({\frac{\displaystyle 1}{\displaystyle 2 }\sqrt{3 }}\)

g) \({\frac{\displaystyle 2}{\displaystyle 3 }\sqrt{6 }}\)

h) \({-\frac{\displaystyle 4}{\displaystyle 3 }\sqrt{3 }}\)

i) \({-\frac{\displaystyle 2}{\displaystyle 5 }\sqrt{5 }}\)

j) \({-2^3\sqrt{2}}\)

k) \({-10\sqrt{\frac{\displaystyle 1}{\displaystyle 10 }}}\)

Arată rezolvarea

Folosim formulele

\({a\sqrt{b} = \sqrt{a^2 \cdot b}}\),

\({-a\sqrt{b} = -\sqrt{a^2 \cdot b}}\),

\({\frac{\displaystyle a}{\displaystyle b }\sqrt{c} = \sqrt{\left(\frac{\displaystyle a}{\displaystyle b }\right)^2 \cdot c}}\),

\({-\frac{\displaystyle a}{\displaystyle b }\sqrt{c} = -\sqrt{\left(\frac{\displaystyle a}{\displaystyle b }\right)^2 \cdot c}}\),

unde \({a \ge 0}\), \({b > 0}\) și \({c \ge 0}\).

⁇ De ce am pus condiția \({b > 0}\) și nu a rămas \({b \ge 0}\)? Altfel spus, de ce \({b }\) nu poate fi 0?

Numărul \({b }\) nu poate fi 0 pentru că el este numitorul fracției \({\frac{\displaystyle a}{\displaystyle b } }\) și numitorul unei fracții nu poate fi egal cu 0.

a) \({\textcolor{deeppink}{7}\sqrt{2}=\sqrt{\textcolor{deeppink}{7}^\textcolor{#6495ed}{\textbf{2}} \cdot 2}=\sqrt{49 \cdot 2}=\sqrt{98}}\)

b) \({\textcolor{deeppink}{3}\sqrt{3}=\sqrt{\textcolor{deeppink}{3}^\textcolor{#6495ed}{\textbf{2}} \cdot 3}=\sqrt{9 \cdot 3}=\sqrt{27}}\)

c) \({\textcolor{deeppink}{6}\sqrt{2}=\sqrt{\textcolor{deeppink}{6}^\textcolor{#6495ed}{\textbf{2}} \cdot 2}=\sqrt{36 \cdot 2}=\sqrt{72}}\)

d) \({\textcolor{deeppink}{3^2}\sqrt{2}=\sqrt{\textcolor{deeppink}{(3^2)}^\textcolor{#6495ed}{\textbf{2}} \cdot 2}=\sqrt{9^2 \cdot 2}=\sqrt{81 \cdot 2}=\sqrt{162}}\)

e) \({\textcolor{deeppink}{2^2}\sqrt{5}=\sqrt{\textcolor{deeppink}{(2^2)}^\textcolor{#6495ed}{\textbf{2}} \cdot 5}=\sqrt{4^2 \cdot 5}=\sqrt{16 \cdot 5}=\sqrt{80}}\)

f) \({\textcolor{deeppink}{\frac{\displaystyle 1}{\displaystyle 2}}\sqrt{3}=\sqrt{\textcolor{deeppink}{\left(\frac{\displaystyle 1}{\displaystyle 2 }\right)}^\textcolor{#6495ed}{\textbf{2}} \cdot 3} }\)

\({\textcolor{white}{\frac{\displaystyle 1}{\displaystyle 2}\sqrt{3}} =\sqrt{\frac{\displaystyle 1^2}{\displaystyle 2^2 } \cdot 3}}\)

\({\textcolor{white}{\frac{\displaystyle 1}{\displaystyle 2}\sqrt{3}} =\sqrt{\frac{\displaystyle 1}{\displaystyle 4 } \cdot 3}}\)

\({\textcolor{white}{\frac{\displaystyle 1}{\displaystyle 2}\sqrt{3}} =\sqrt{\frac{\displaystyle 1 \cdot 3}{\displaystyle 4 }}}\)

\({\textcolor{white}{\frac{\displaystyle 1}{\displaystyle 2}\sqrt{3}} =\sqrt{\frac{\displaystyle 3}{\displaystyle 4 }}}\)

g) \({\textcolor{deeppink}{\frac{\displaystyle 2}{\displaystyle 3}}\sqrt{6}=\sqrt{\textcolor{deeppink}{\left(\frac{\displaystyle 2}{\displaystyle 3 }\right)}^\textcolor{#6495ed}{\textbf{2}} \cdot 6} }\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 3}\sqrt{6}} =\sqrt{\frac{\displaystyle 2^2}{\displaystyle 3^2 } \cdot 6}}\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 3}\sqrt{6}} =\sqrt{\frac{\displaystyle 4}{\displaystyle \cancel{9}_3 } \cdot \cancel{6}^2}}\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 3}\sqrt{6}} =\sqrt{\frac{\displaystyle 4 \cdot 2}{\displaystyle 3 }}}\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 3}\sqrt{6}} =\sqrt{\frac{\displaystyle 8}{\displaystyle 3 }}}\)

h) \({-\textcolor{deeppink}{\frac{\displaystyle 4}{\displaystyle 3}}\sqrt{3}=-\sqrt{\textcolor{deeppink}{\left(\frac{\displaystyle 4}{\displaystyle 3 }\right)}^\textcolor{#6495ed}{\textbf{2}} \cdot 3} }\)

Semnul „-” (minus) rămâne în fața radicalului.

\({\textcolor{white}{\frac{\displaystyle 4}{\displaystyle 3}\sqrt{3}} =-\sqrt{\frac{\displaystyle 4^2}{\displaystyle 3^\cancel{2} } \cdot \cancel{3}}}\)

\({\textcolor{white}{\frac{\displaystyle 4}{\displaystyle 3}\sqrt{3}} =-\sqrt{\frac{\displaystyle 4^2}{\displaystyle 3 }}}\)

\({\textcolor{white}{\frac{\displaystyle 4}{\displaystyle 3}\sqrt{6}} =-\sqrt{\frac{\displaystyle 16}{\displaystyle 3}}}\)

i) \({-\textcolor{deeppink}{\frac{\displaystyle 2}{\displaystyle 5}}\sqrt{5}=-\sqrt{\textcolor{deeppink}{\left(\frac{\displaystyle 2}{\displaystyle 5 }\right)}^\textcolor{#6495ed}{\textbf{2}} \cdot 5} }\)

Semnul „-” (minus) rămâne în fața radicalului.

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 5}\sqrt{5}} =-\sqrt{\frac{\displaystyle 2^2}{\displaystyle 5^\cancel{2} } \cdot \cancel{5}}}\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 5}\sqrt{5}} =-\sqrt{\frac{\displaystyle 2^2}{\displaystyle 5 }}}\)

\({\textcolor{white}{\frac{\displaystyle 2}{\displaystyle 5}\sqrt{5}} =-\sqrt{\frac{\displaystyle 4}{\displaystyle 5}}}\)

j) \({-\textcolor{deeppink}{2^3}\sqrt{2}=-\sqrt{\textcolor{deeppink}{(2^3)}^\textcolor{#6495ed}{\textbf{2}} \cdot 2} }\)

Semnul „-” (minus) rămâne în fața radicalului.

\({\textcolor{white}{2^3\sqrt{2}} =-\sqrt{2^{3 \; \cdot \; 2} \; \cdot \; 2^1}}\)

\({\textcolor{white}{2^3\sqrt{2}} =-\sqrt{2^{6} \cdot 2^1}}\)

\({\textcolor{white}{2^3\sqrt{2}} =-\sqrt{2^{6 \; + \; 1}}}\)

\({\textcolor{white}{2^3\sqrt{2}} =-\sqrt{2^{7}}}\)

\({\textcolor{white}{2^3\sqrt{2}} =-\sqrt{128}}\)

k) \({-\textcolor{deeppink}{10}\sqrt{\frac{\displaystyle 1}{\displaystyle 10 }}=-\sqrt{\textcolor{deeppink}{10}^\textcolor{#6495ed}{\textbf{2}} \cdot \frac{\displaystyle 1}{\displaystyle 10 }} }\)

Semnul „-” (minus) rămâne în fața radicalului.

\({\textcolor{white}{10\sqrt{\frac{\displaystyle 1}{\displaystyle 10 }}}=-\sqrt{10^{\cancel{2}^1} \cdot \frac{\displaystyle 1}{\displaystyle \cancel{10} } } }\)

\({\textcolor{white}{10\sqrt{\frac{\displaystyle 1}{\displaystyle 10} }}=-\sqrt{10}}\)

B. Adevărat sau fals? Completați cu A pentru afirmațiile adevărate sau cu F pentru afirmațiile false:

a) \({3\sqrt{16}=\sqrt{144}}\)

b) \({5\sqrt{6}=\sqrt{150}}\)

c) \({2\sqrt{22}=\sqrt{2^2 \cdot 22}}\)

d) \({2\sqrt{10}=\sqrt{20}}\)

e) \({-4\sqrt{3}=\sqrt{(-4)^2 \cdot 3}}\)

f) \({-2^3\sqrt{5}=-\sqrt{(2^3)^2 \cdot 5}=-\sqrt{2^9 \cdot 5}}\)

g) \({-4\sqrt{10}=\sqrt{160}}\)

h) \({-3\sqrt{21}=-\sqrt{3^2 \cdot 21}}\)

i) \({\frac{\displaystyle 2}{\displaystyle 3}\sqrt{\frac{\displaystyle 3}{\displaystyle 2}}=\sqrt{\left(\frac{\displaystyle 2}{\displaystyle 3}\right)^2 \cdot \frac{\displaystyle 3}{\displaystyle 2}}=\sqrt{\frac{\displaystyle 2^\cancel{2}}{\displaystyle 3^\cancel{2}} \cdot \frac{\displaystyle \cancel{3}}{\displaystyle \cancel{2}}}=\sqrt{\frac{\displaystyle 2}{\displaystyle 3}}}\)

j) \({-\frac{\displaystyle 3}{\displaystyle 5}\sqrt{5}=-\sqrt{\left(\frac{\displaystyle 3}{\displaystyle 5}\right)^2 \cdot 5}=-\sqrt{\frac{\displaystyle 3^2}{\displaystyle 5^\cancel{2}} \cdot \cancel{5}}=-\sqrt{\frac{\displaystyle 9}{\displaystyle 5}}}\)

k) \({-10^3\sqrt{\frac{\displaystyle 1}{\displaystyle 10}}=-\sqrt{(10^3)^{\cancel{2}^1}\cdot \frac{\displaystyle 1}{\displaystyle \cancel{10}}}=-\sqrt{10^3}=-\sqrt{1000}}\)

Arată rezolvarea

Folosim formulele

\({a\sqrt{b} = \sqrt{a^2 \cdot b}}\),

\({-a\sqrt{b} = -\sqrt{a^2 \cdot b}}\),

\({\frac{\displaystyle a}{\displaystyle b }\sqrt{c} = \sqrt{\left(\frac{\displaystyle a}{\displaystyle b }\right)^2 \cdot c}}\),

\({-\frac{\displaystyle a}{\displaystyle b }\sqrt{c} = -\sqrt{\left(\frac{\displaystyle a}{\displaystyle b }\right)^2 \cdot c}}\),

unde \({a \ge 0}\), \({b > 0}\) și \({c \ge 0}\).

‼ Semnul - (minus) rămâne în fața radicalului.

a) \({3\sqrt{16}=\sqrt{144}}\) A (adevărat)

\({\textcolor{deeppink}{3}\sqrt{16}=\sqrt{\textcolor{deeppink}{3}^\textcolor{#6495ed}{\textbf{2}} \cdot 16}=\sqrt{9 \cdot 16}=\sqrt{144}}\)

b) \({5\sqrt{6}=\sqrt{150}}\) A (adevărat)

\({\textcolor{deeppink}{5}\sqrt{6}=\sqrt{\textcolor{deeppink}{5}^\textcolor{#6495ed}{\textbf{2}} \cdot 6}=\sqrt{25 \cdot 6}=\sqrt{150}}\)

c) \({2\sqrt{22}=\sqrt{2^2\cdot 22 }}\) A (adevărat)

\({\textcolor{deeppink}{2}\sqrt{22}=\sqrt{\textcolor{deeppink}{2}^\textcolor{#6495ed}{\textbf{2}} \cdot 22}\textcolor{#778899}{=\sqrt{4 \cdot 22}=\sqrt{88}}}\)

d) \({2\sqrt{10}=\sqrt{20}}\) F (fals)

\({\textcolor{deeppink}{2}\sqrt{10}=\sqrt{\textcolor{deeppink}{2}^\textcolor{#6495ed}{\textbf{2}} \cdot 10}=\sqrt{4 \cdot 10}=\sqrt{40}}\)

e) \({-4\sqrt{3}=\sqrt{(-4)^2 \cdot 3}}\) F (fals)

\({-\textcolor{deeppink}{4}\sqrt{3}=-\sqrt{\textcolor{deeppink}{4}^\textcolor{#6495ed}{\textbf{2}} \cdot 3}\textcolor{#778899}{=-\sqrt{16 \cdot 3}=-\sqrt{48}}}\)

f) \({-2^3\sqrt{5}=-\sqrt{(2^3)^2 \cdot 5}=-\sqrt{2^9 \cdot 5}}\) F (fals)

\({-\textcolor{deeppink}{2^3}\sqrt{5}=-\sqrt{\textcolor{deeppink}{(2^3)}^\textcolor{#6495ed}{\textbf{2}} \cdot 5}=-\sqrt{2^{3 \; \cdot \; 2} \cdot 5}=-\sqrt{2^6 \cdot 5}\textcolor{#778899}{=-\sqrt{64 \cdot 5}=-\sqrt{320}}}\)

\({(a^m)^n=a^{m \; \cdot \; n}}\)

g) \({-4\sqrt{10}=\sqrt{160}}\) F (fals)

\({-\textcolor{deeppink}{4}\sqrt{10}=-\sqrt{\textcolor{deeppink}{4}^\textcolor{#6495ed}{\textbf{2}} \cdot 10}=-\sqrt{16 \cdot 10}=-\sqrt{160}}\)

h) \({-3\sqrt{21}=-\sqrt{3^2 \cdot 21}}\) A (adevărat)

\({-\textcolor{deeppink}{3}\sqrt{21}=-\sqrt{\textcolor{deeppink}{3}^\textcolor{#6495ed}{\textbf{2}} \cdot 21}\textcolor{#778899}{=-\sqrt{9 \cdot 21}=-\sqrt{189}}}\)

\({\left(\frac{\displaystyle a}{\displaystyle b}\right)^m=\frac{\displaystyle a^m}{\displaystyle b^m}}\)

k) \({-10^3\sqrt{\frac{\displaystyle 1}{\displaystyle 10}}=-\sqrt{(10^3)^{\cancel{2}^1}\cdot \frac{\displaystyle 1}{\displaystyle \cancel{10}}}=-\sqrt{10^3}=-\sqrt{1000}}\) F (fals)

\({-\textcolor{deeppink}{10^3}\sqrt{\frac{\displaystyle 1}{\displaystyle 10}}=-\sqrt{\textcolor{deeppink}{(10^3)}^\textcolor{#6495ed}{\textbf{2}} \cdot \frac{\displaystyle 1}{\displaystyle 10}}=-\sqrt{10^{3 \; \cdot \; 2} \cdot \frac{\displaystyle 1}{\displaystyle 10}}=-\sqrt{10^{\cancel{6}^5} \cdot \frac{\displaystyle 1}{\displaystyle \cancel{10}}}=-\sqrt{10^5}=-\sqrt{100000}}\)

C. Completați cu semnul <, > sau =, astfel încât să obțineți afirmații adevărate:

a) \({3\sqrt{2}}\) \({2\sqrt{3}}\)

b) \({4\sqrt{8}}\) \({5\sqrt{6}}\)

c) \({-2\sqrt{6}}\) \({-3\sqrt{5}}\)

d) \({-4\sqrt{10}}\) \({2\sqrt{11}}\)

e) \({10}\) \({2\sqrt{10}}\)

f) \({\sqrt{32}}\) \({4\sqrt{2}}\)

g) \({-3\sqrt{3}}\) \({-8\sqrt{3}}\)

Arată rezolvarea

Fie \({x}\) și \({y}\) numere reale pozitive.

dacă \({x < y}\), atunci \({\sqrt{x} < \sqrt{y}}\);
dacă \({\sqrt{x} < \sqrt{y}}\), atunci \({x < y}\);
dacă vrem să comparăm numere de forma \({a\sqrt{x}, \text{cu} \; a \ge 0}\), atunci mai întâi introducem factorii sub radical, apoi comparăm numerele de sub radical.

a) \({3\sqrt{2}}\) > \({2\sqrt{3}}\)

introducem factorii sub radical
\({\textcolor{deeppink}{3}\sqrt{2}=\sqrt{\textcolor{deeppink}{3}^\textcolor{#6495ed}{\textbf{2}} \cdot 2}=\sqrt{9 \cdot 2}=\sqrt{18}}\)
\({\textcolor{deeppink}{2}\sqrt{3}=\sqrt{\textcolor{deeppink}{2}^\textcolor{#6495ed}{\textbf{2}} \cdot 3}=\sqrt{4 \cdot 3}=\sqrt{12}}\)
deoarece \({18 > 12}\), rezultă că \({\sqrt{18} > \sqrt{12}}\)

b) \({4\sqrt{8}}\) < \({5\sqrt{6}}\)

introducem factorii sub radical
\({\textcolor{deeppink}{4}\sqrt{8}=\sqrt{\textcolor{deeppink}{4}^\textcolor{#6495ed}{\textbf{2}} \cdot 8}=\sqrt{16 \cdot 8}=\sqrt{128}}\)
\({\textcolor{deeppink}{5}\sqrt{6}=\sqrt{\textcolor{deeppink}{5}^\textcolor{#6495ed}{\textbf{2}} \cdot 6}=\sqrt{25 \cdot 6}=\sqrt{150}}\)
deoarece \({128 < 150}\), rezultă că \({\sqrt{128} < \sqrt{150}}\)

c) \({-2\sqrt{6}}\) > \({-3\sqrt{5}}\)

introducem factorii sub radical
\({-\textcolor{deeppink}{2}\sqrt{6}=-\sqrt{\textcolor{deeppink}{2}^\textcolor{#6495ed}{\textbf{2}} \cdot 6}=-\sqrt{4 \cdot 6}=-\sqrt{24}}\)
\({-\textcolor{deeppink}{3}\sqrt{5}=-\sqrt{\textcolor{deeppink}{3}^\textcolor{#6495ed}{\textbf{2}} \cdot 5}=-\sqrt{9 \cdot 5}=-\sqrt{45}}\)
deoarece \({24 < 45}\), rezultă că \({\sqrt{24} < \sqrt{45} \;\; \mid \;\; \cdot \; (-1)}\)
rezultă că \({-\sqrt{24} > -\sqrt{45} }\)
când înmulțim o inegalitate cu un număr negativ, se schimbă sensul inegalității; în cazul nostru, < devine >

d) \({-4\sqrt{10}}\) < \({2\sqrt{11}}\)

orice număr negativ este mai mic decât orice număr pozitiv
\({-4\sqrt{10}}\) este număr negativ
\({2\sqrt{11}}\) este număr pozitiv
rezultă că \({-4\sqrt{10} < 2\sqrt{11}}\)

e) \({10}\) > \({2\sqrt{10}}\)

introducem factorii sub radical
avem \({10=\sqrt{100}}\)
\({\textcolor{deeppink}{2}\sqrt{10}=\sqrt{\textcolor{deeppink}{2}^\textcolor{#6495ed}{\textbf{2}} \cdot 10}=\sqrt{4 \cdot 10}=\sqrt{40}}\)
deoarece \({100 > 40}\), rezultă că \({\sqrt{100} > \sqrt{40}}\), adică \({10 > 2\sqrt{10}}\)

f) \({\sqrt{32}}\) = \({4\sqrt{2}}\)

introducem factorii sub radical
\({\textcolor{deeppink}{4}\sqrt{2}=\sqrt{\textcolor{deeppink}{4}^\textcolor{#6495ed}{\textbf{2}} \cdot 2}=\sqrt{16 \cdot 2}=\sqrt{32}}\)
deoarece \({32=32}\), rezultă că \({\sqrt{32} =4\sqrt{2}}\)

g) \({-3\sqrt{3}}\) > \({-8\sqrt{3}}\)

observăm că avem același radical, deci comparăm factorii care sunt în fața radicalului
deoarece \({-3 > -8}\), rezultă că \({-3\sqrt{3} > -8\sqrt{3}}\)

Exersează 1 | Exersează 2

⮜ Scoaterea factorilor de sub radical

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