a ) Đặt \(\sqrt{x+1}=a\Rightarrow x+1=a^2\Rightarrow x=a^2-1\)
Ta có : \(x^2+x+12\sqrt{x+1}=36\)
\(\Leftrightarrow x\left(x+1\right)+12a=36\)
\(\Leftrightarrow a^2\left(a^2-1\right)+12a-36=0\)
\(\Leftrightarrow a^4-a^2+12a-36=0\)
\(\Leftrightarrow a^3\left(a-2\right)+2a^2\left(a-2\right)+3a\left(a-2\right)+18\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2+3a+18\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left[a^2\left(a+3\right)-a\left(a+3\right)+6\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\left(a^2-a+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+1}=-3\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x+1=4\Leftrightarrow x=3\)
Vậy ...
b ) \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow\left(x^4-8x^2+16\right)+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(4-x^2=a\) , ta có :
\(a^2+x-4=0\) \(\Rightarrow x=4-a^2\)
Ta có : x = \(4-a^2;a=4-x^2\)
\(\Leftrightarrow x-a=x^2-a^2\)
\(\Leftrightarrow\left(x-a\right)\left(1-x-a\right)=0\)
\(\Leftrightarrow\left(x-4+x^2\right)\left(1-x-4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2+x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow...\)
