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\(6x^2+10x-92+\sqrt{\left(x+70\right)\left(2x^2+4x+16\right)}=0\)(*)

Đặt \(a=\sqrt{x+70};\sqrt{2x^2+4x+16}=b\), (*) trở thành:

\(6x^2+10x-92+ab=0\)

\(\Leftrightarrow6x^2+12x+48-2x-140+ab=0\)

\(\Leftrightarrow3b^2-2a^2+ab=0\)

\(\Leftrightarrow3b^2+3ab-2ab-2a^2=0\)

\(\Leftrightarrow3b\left(a+b\right)-2a\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(3b-2a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\3b=2a\end{matrix}\right.\)

Tới đây dễ rồi UwU

Chú ý rằng \(3\left(2x^2+4x+16\right)-2\left(x+70\right)=6x^2+10x-92\)

ĐKXĐ: \(x\ge-70\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+70}=a\ge0\\\sqrt{2x^2+4x+16}=b>0\end{matrix}\right.\) \(\Rightarrow a+b>0\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành: \(3a^2+ab-2b^2=0\Leftrightarrow\left(3a-2b\right)\left(a+b\right)=0\)

\(\Leftrightarrow3a-2b=0\Rightarrow3a=2b\)

\(\Rightarrow3\sqrt{x+70}=2\sqrt{2x^2+4x+16}\Leftrightarrow9\left(x+70\right)=4\left(2x^2+4x+16\right)\)

\(\Leftrightarrow8x^2+7x-566=0\Rightarrow\left[{}\begin{matrix}x=...\\x=...\end{matrix}\right.\)

Đặt \(\left(x^2-x+1\right)^2=a;x^2=b\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a^2-10ab+9b^2=0\\ \Leftrightarrow a^2-9ab-ab+9b^2=0\\ \Leftrightarrow\left(a-b\right)\left(a-9b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=9b\end{matrix}\right.\\ \forall a=b\Leftrightarrow\left(x^2-x+1\right)^2-x^2=0\\ \Leftrightarrow\left(x^2-2x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow x=1\\ \forall a=9b\Leftrightarrow\left(x^2-x+1\right)^2-9x^2=0\\ \Leftrightarrow\left(x^2-4x+1\right)\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Lời giải:

1. 

PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$

$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)

$\Leftrightarrow (t-4)(t+6)=0$

$\Rightarrow t-4=0$ hoặc $t+6=0$

Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$

$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$

Nếu $t+6=0$

$\Leftrightarrow x^2+5x+6=0$

$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$

2.

PT $\Leftrightarrow (x^2-4x+1)^2+2(x^2-4x+1)-3=0$

$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)

$\Leftrightarrow (t-1)(t+3)=0$

$\Rightarrow t-1=0$ hoặc $t+3=0$

Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$

$\Rightarrow x=0$ hoặc $x=4$

Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$

$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$

b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)