Đề bài yêu cầu j vậy?

mũ 2 tui ko bt

\(3.x^{4+5}.x^2.y^{2+2}.y^{4+2}.y^2\)

NTN đúng ko??

a) \(x:\left(\frac{-1}{2}\right)^3=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2}.\left(\frac{-1}{2}\right)^3\)

\(\Rightarrow x=\left(\frac{-1}{2}\right)^4=\frac{1}{16}\)

b) \(\left(\frac{3}{4}\right)^5.x=\left(\frac{3}{4}\right)^7\)

\(\Rightarrow x=\left(\frac{3}{4}\right)^7:\left(\frac{3}{4}\right)^5\)

\(\Rightarrow x=\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Cảm ơn bn 

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x^3}{125}=\dfrac{y^3}{64}=\dfrac{z^3}{8}=\dfrac{x^3-y^3+z^3}{125-64+8}=\dfrac{69}{69}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt[3]{125}=5\\y=\sqrt[3]{64}=4\\z=\sqrt[3]{8}=2\end{matrix}\right.\)

x bằng bao nhiêu cậu ơi ?

x=\(\frac{y}{2}\)=\(\frac{z}{3}\)

3.

a) \(\left(x-1\right)^3=125\)

=> \(\left(x-1\right)^3=5^3\)

=> \(x-1=5\)

=> \(x=5+1\)

=> \(x=6\)

Vậy \(x=6.\)

b) \(2^{x+2}-2^x=96\)

=> \(2^x.\left(2^2-1\right)=96\)

=> \(2^x.3=96\)

=> \(2^x=96:3\)

=> \(2^x=32\)

=> \(2^x=2^5\)

=> \(x=5\)

Vậy \(x=5.\)

c) \(\left(2x+1\right)^3=343\)

=> \(\left(2x+1\right)^3=7^3\)

=> \(2x+1=7\)

=> \(2x=7-1\)

=> \(2x=6\)

=> \(x=6:2\)

=> \(x=3\)

Vậy \(x=3.\)

Chúc bạn học tốt!

Giúp mk với nha các bạn