Question

[grgerger]

giải đc cho 10k

Answer 1

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Bài 2 :

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

\(\Rightarrow\orbr{\begin{cases}\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\\left(5x+1\right)^2=\left(\frac{-6}{7}\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{4}{9}\right)^3\)

\(\Rightarrow x-\frac{2}{9}=\frac{4}{9}\)

\(\Rightarrow x=\frac{2}{3}\)

c) ( x - 2 )³ = 64

=> ( x - 2 )³ = 4³

=> x - 2 = 4

=> x = 6

d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\le0\)

Vì \(\hept{\begin{cases}\left(x-3,5\right)^2\ge0\forall x\\\left(y-\frac{1}{10}\right)^4\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-3,5\right)^2=0\\\left(y-\frac{1}{10}\right)^4=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-3,5=0\\y-\frac{1}{10}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3,5\\y=\frac{1}{10}\end{cases}}\)

Answer 2

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Bài 3 :

a) Ta có : 2²⁴  = ( 2³ )⁸ = 8⁸ < 9⁸ = ( 3² )⁸ = 3¹⁶

=> 2²⁴ < 3¹⁶

b) Ta có : ( -5 )⁹ < 0 < ( -2 )¹⁸

=> ( -5 )⁹ < ( -2 )¹⁸