Question

Решите Алгебра Срочно Пожалуйста

Answer 1

1) $\frac{1}{6}\sqrt{144} =\frac{1}{6}*12=\frac{12}{6}=2$

2) $2,1+1,3*\sqrt{\frac{81}{169}}=2,1 +1,3*\frac{9}{13}=2,1+\frac{13}{10}*\frac{9}{13}=2,1+\frac{13*9}{10*13}=\frac{21}{10}+\frac{9}{10}=\frac{30}{10}=3$

3) $(0,4\sqrt{5} )^{2}=0,4^{2}*5=(\frac{4}{10})^{2}*5=\frac{16}{100}*\frac{5}{1}=\frac{16*5}{100*1}=\frac{16}{25}=\frac{4}{5}=0,8$

4) $\sqrt{225*0,04}=15*0,2=3$

5) $\sqrt{28}*\sqrt{63}=\sqrt{4*7}*\sqrt{9*7}=2\sqrt{7}*3\sqrt{7}=6(\sqrt{7} )^{2}=6*7=42$

6) $\frac{\sqrt{48}}{\sqrt{3}}=\frac{\sqrt{16*3}}{\sqrt{3}}=\frac{4\sqrt{3}}{\sqrt{3}}=4$

7) $\sqrt{2^{6}*7^{2}}=2^{3}*7=8*7=56$

8) $x^{2}=0,09\\x=+-\sqrt{0,09}\\x=+-0,3$

9) $x^{2}=92\\x=+-\sqrt{92}\\x=+-\sqrt{4*23}\\x=+-2\sqrt{23}$

10) $\frac{1}{7}x^{2}\sqrt{49x^{6}}=\frac{1}{7}x^{2}*7x^3=\frac{1*7}{7*1}x^{2+3}=x^{5}$

11) $-5y^{6}\sqrt{\frac{1}{81y^{10}}}=-5y^{6}*\frac{1}{9y^{5}}=-\frac{5y^{6}*1}{1*9y^{5}}=-\frac{5y}{9}$