a) Ta có: \(\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=\frac{7}{2}\)
\(\Leftrightarrow\left(8x+7\right)^2\cdot2\left(4x+3\right)\cdot8\left(x+1\right)=16\cdot\frac{7}{2}\)
\(\Leftrightarrow\left(8x+7\right)^2\left(8x+6\right)\left(8x+8\right)=56\)
Đặt \(8x+7=a\) khi đó:
\(a^2\left(a-1\right)\left(a+1\right)=56\)
\(\Leftrightarrow a^2\left(a^2-1\right)=56\)
\(\Leftrightarrow a^4-a^2-56=0\)
\(\Leftrightarrow\left(a^2-8\right)\left(a^2+7\right)=0\)
\(\Leftrightarrow a^2-8=0\Leftrightarrow\left(8x+7\right)^2-8=0\)
\(\Leftrightarrow\left(8x+7\right)^2=8\Leftrightarrow\orbr{\begin{cases}8x+7=2\sqrt{2}\\8x+7=-2\sqrt{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}8x=2\sqrt{2}-7\\8x=-2\sqrt{2}-7\end{cases}}\Rightarrow x=\frac{\pm2\sqrt{2}-7}{8}\)
b) Ta có: \(x^2+5y^2-4xy+10x-22y+\left|x+y+z\right|+26=0\)
\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(10x-20y\right)+25+y^2-2y+1+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y\right)^2+10\left(x-2y\right)+25+\left(y-1\right)^2+\left|x+y+z\right|=0\)
\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-2y+5\right)^2=0\\\left(y-1\right)^2=0\\\left|x+y+z\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\\z=2\end{cases}}\)
Vậy x = -3 , y = 1 , z = 2
a, Phương trình \(\Leftrightarrow\left(8x+7\right)^2\left(8x+6\right)\left(8x+8\right)=56\)
\(\Leftrightarrow\left(64x^2+112x+48\right)\left(64x^2+112x+49\right)=56\)
Đặt \(a=64x^2+112x+\frac{97}{2}\), ta có phương trình: ...
