Các bạn giúp mình với ạ! TT

\(.a.\)

\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}=0\)

\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)

\(\Leftrightarrow\left[\begin{matrix}\left(x-7\right)^{x+1}=0\\\left[1-\left(x-7\right)^{10}\right]=0\end{matrix}\right.\)

+ Nếu \(\left(x-7\right)^{x+1}=0\)

\(\Rightarrow x-7=0\)

\(\Rightarrow x=0+7\)

\(\Rightarrow x=7\)

+ Nếu \(1-\left(x-7\right)^{10}=0\)

\(\Rightarrow\left(x-7\right)^{10}=1\)

\(\Rightarrow\left(x-7\right)^{10}=\left(\pm1\right)^{10}\)

\(\Rightarrow\left[\begin{matrix}x-7=1\\x-7=-1\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=1+7\\x=-1+7\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=8\\x=6\end{matrix}\right.\)

Vậy : \(x\in\left\{6;7;8\right\}\)

c)\(\left|2x+3\right|=x+2\)

Đk:\(x+2\ge0\Rightarrow x\ge-2\)

TH1:2x+3=x+2

\(\Rightarrow2x-x=2-3\)

\(\Rightarrow x=-1\)(Thỏa mãn đk )

TH2:2x+3=-x-2

\(\Rightarrow2x+x=-2+3\)

\(\Rightarrow3x=1\)

\(\Rightarrow x=\frac{1}{3}\)(Thỏa mãn đk)

Vậy x=-1 hoặc x=1/3

Từ \(\frac{5x-1}{3}=\frac{7y-6}{5}\) Áp dụng TC DTSBN ta có :

\(\frac{5x-1}{3}=\frac{7y-6}{5}=\frac{\left(5x-1\right)+\left(7y-6\right)}{3+5}=\frac{5x+7y-7}{8}=\frac{5x+7y-7}{4x}\)

\(\Rightarrow4x=8\Rightarrow x=2\)

\(\Rightarrow\frac{5.2-1}{3}=\frac{7y-6}{5}\)

\(\Leftrightarrow\frac{7y-6}{5}=3\)

\(\Rightarrow y=3\)

Vậy \(x=2;y=3\)

\(\frac{5x-1}{3}=\frac{7y-6}{5}\Rightarrow5\left(5x-1\right)=3\left(7y-6\right)\Rightarrow25x-5=21y-18\)

\(\Rightarrow21y=25x+13\Rightarrow7y=\frac{25x+13}{3}\)

Xét : \(\frac{5x+7y-7}{4x}=\frac{5x+\frac{25x+13}{3}-7}{4x}=\frac{10x-2}{3x}\)

\(\Rightarrow3x\left(5x-1\right)=3\left(10x-2\right)\Rightarrow15x^2-33x+6=0\)

\(\Rightarrow3\left(x-2\right)\left(5x-1\right)=0\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{5}\end{cases}}\)

Với x=2 , ta có : y=3

Với x =\(\frac{1}{5}\), ta có : y= \(\frac{6}{7}\)