a) \(7x-8=4x+7\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

b) \(\frac{5x-4}{12}=\frac{16x+1}{7}\)

\(\Leftrightarrow35x-28=192x+12\)

\(\Leftrightarrow157x=-40\Leftrightarrow x=\frac{-40}{157}\)

c)\(ĐKXĐ:x\ne\pm2\)

 \(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)

\(\Rightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow\frac{y^2+3y+2-5y+10}{y^2-4}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow y^2-2y+12=12+y^2-4\)

\(\Rightarrow-2y=-4\Leftrightarrow y=2\left(ktm\right)\)

Vậy pt vô nghiệm

\(a,7x-8=4x+7\)

\(7x-4x=7+8\)

\(3x=15\Leftrightarrow5\)

\(b,\frac{5x-4}{12}=\frac{16x+1}{7}\)

\(\left(5x-4\right)7=\left(16x+1\right)12\)

\(35x-28=192x+12\)

\(35x-192x=12+28\)

\(-157x=40\Leftrightarrow x=-\frac{40}{157}\)

\(\frac{y-1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)

 \(\frac{\left(y-1\right)\left(y+2\right)}{y^2-4}-\frac{5\left(y-2\right)}{y^2-4}=\frac{12}{y^2-4}+\frac{y^2-4}{y^2-4}\)

\(\frac{y^2+y-2-5y+10}{y^2-4}=\frac{y^2+8}{y^2-4}\)

\(y^2-4y-8=y^2+8\)

 \(y^2-4y-8-y^2-8=0\)

 \(-4y-16=0\)

\(\Rightarrow y=-4\)

            Vậy y=-4

\(\Leftrightarrow\frac{y-1}{y-2}-\frac{5}{y+2}=\frac{12}{\left(y-2\right)\left(y+2\right)}+1\)

\(\Leftrightarrow\frac{\left(y-1\right)\left(y+2\right)-5\left(y-2\right)-12+1\left(y-2\right)\left(y+2\right)}{\left(y-2\right)\left(y+2\right)}=0\)

\(\Leftrightarrow\frac{y^2+2y-y-2-5y+10-12+y^2+2y-2y-4}{\left(y-2\right)\left(y+2\right)}\)

Rồi bạn làm tiếp nha

\(x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}\ge2\sqrt{\frac{x^2}{x^2}}+2\sqrt{\frac{y^2}{y^2}}=4\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x^2=\frac{1}{x^2}\\y^2=\frac{1}{y^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\end{matrix}\right.\)

Thay vào pt đầu thì chỉ \(x=y=1\) phù hợp

Điều kiện xác định: \(x\ne-1;y\ne1\)

\(\hept{\begin{cases}\frac{x^2}{x+1}+\frac{y^2}{y-1}=4\left(1\right)\\\frac{x+2}{x+1}+\frac{y-2}{y-1}=y-x\left(2\right)\end{cases}}\)

Từ pt (2), ta có: \(\frac{x+2}{x+1}+\frac{y-2}{y-1}=y-x\)

\(\Leftrightarrow1+\frac{1}{x+1}+1-\frac{1}{y-1}-y+x=0\)

\(\Leftrightarrow x+1+\frac{1}{x+1}-\left(y-1+\frac{1}{y-1}\right)=0\)

\(\Leftrightarrow x+1+\frac{1}{x+1}=y-1+\frac{1}{y-1}\)

\(\Leftrightarrow\frac{x^2+2x+2}{x+1}=\frac{y^2-2y+2}{y-1}\)

\(\Leftrightarrow\frac{x^2}{x+1}+\frac{2\left(x+1\right)}{x+1}=\frac{y^2}{y+1}-\frac{2\left(y-1\right)}{y-1}\)

\(\Leftrightarrow\frac{x^2}{x+1}+2=\frac{y^2}{y-1}-2\)

\(\Leftrightarrow\frac{x^2}{x+1}+4-\frac{y^2}{y-1}=0\)(*)

Thay (1) vào (*), ta được: \(\frac{x^2}{x+1}+\frac{x^2}{x+1}+\frac{y^2}{y-1}-\frac{y^2}{y-1}=0\)

\(\Leftrightarrow\frac{2x^2}{x+1}=0\)

\(\Leftrightarrow2x^2=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

Thay x = 0 vào pt (1), ta được: \(\frac{y^2}{y-1}=4\) \(\Leftrightarrow\left(y-2\right)^2=0\) \(\Leftrightarrow y=2\left(tm\right)\)

Vậy: Hệ có nghiệm duy nhất thỏa mãn: \(\left(0;2\right)\)

=.= hk tốt!!