ĐK x>= -1 

Đặt \(\sqrt{x+1}=a\Rightarrow x=a^2-1\)

pt <=> \(\left(a^2-1\right)^2+a^2-1+12a=36\Leftrightarrow a^4-a^2+12a-36=0\)

<=>  \(\left(a-2\right)\left(a+3\right)\left(a^2-a+6\right)=0\)

<=> a = 2 hoặc a = -3 hoặc \(a^2-a+6=0\)

(+) a = 2 => x = \(3\)

(+) a = -3 ( loại vì \(\sqrt{x+1}\ge0\) )

(+) \(a^2-a+6=a^2-a+\frac{1}{4}+\frac{23}{4}=\left(a-\frac{1}{2}\right)^2+\frac{23}{4}>0\) => pt vô nghệm 

Vậy x = 3 là nghiệm của pt 

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 = \(4-a^2;a=4-x^2\)

\(\Leftrightarrow x-a=x^2-a^2\)

\(\Leftrightarrow\left(x-a\right)\left(1-x-a\right)=0\)

\(\Leftrightarrow\left(x-4+x^2\right)\left(1-x-4+x^2\right)=0\)

\(\Leftrightarrow\left(x^2+x-4\right)\left(x^2+x-3\right)=0\)

\(\Leftrightarrow...\)

phả là 10x chứ

đặt  2 căn đầu bằng a

bình phương a lên

ĐLXĐ:\(x\ge-1\)

\(\sqrt{x^2+4x+12}=2x-4+\sqrt{x+1}\)

\(\Leftrightarrow\left[\sqrt{x^2+4x+12}-\left(6-3x\right)\right]-\left[\sqrt{x+1}-\left(x-2\right)\right]=0\)

\(\Leftrightarrow\frac{x^2+4x+12-36+36x-9x^2}{\sqrt{x^2+4x+12}+2-3x}-\frac{x+1-x^2+4x-4}{\sqrt{x+1}+x+2}=0\)

\(\Leftrightarrow\frac{-8x^2+40x-24}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\frac{8\left(-x^2+5x-3\right)}{\sqrt{x^2+4x+12}+2-3x}-\frac{-x^2+5x-3}{\sqrt{x+1}+x-2}=0\)

\(\Leftrightarrow\left(-x^2+5x-3\right)\left[\frac{8}{\sqrt{x^2+4x+12}+2-3x}-\frac{1}{\sqrt{x+1}+x-2}\right]=0\)

TH1:\(-x^2+5x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

TH2:........ ( chắc vô nghiệm )

phần mẫu phải là \(\sqrt{x^2+4x+12}+6-3x\) chứ :vv Hơi lỗi nhưng cảm ơn nhé !!

\(x^2+4x+12=\left(x+1\right)^2+2\left(x+1\right)+9\)

Đặt \(\sqrt{x+1}=a\ge0\).

PT \(\Leftrightarrow\sqrt{a^4+2a^2+9}=2a^2+a-6\)

1/ ĐKXĐ:...

\(\Leftrightarrow\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-2\sqrt{x+1}+1}=\frac{x+5}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(1-\sqrt{x+1}\right)^2}=\frac{x+5}{2}\)

\(\Leftrightarrow\sqrt{x+1}+1+\left|1-\sqrt{x+1}\right|=\frac{x+5}{2}\)

Nếu \(0\ge x\ge-1\Rightarrow\left|1-\sqrt{x+1}\right|=1-\sqrt{x+1}\)

\(\Rightarrow2=\frac{x+5}{2}\Leftrightarrow x=-1\left(tm\right)\)

Nếu \(x>0\Rightarrow\left|1-\sqrt{x+1}\right|=\sqrt{x+1}-1\)

\(\Rightarrow2\sqrt{x+1}=\frac{x+5}{2}\Leftrightarrow16x+16=x^2+10x+25\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow x=3\left(tm\right)\)

Vậy...

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