a.

ĐKXĐ: \(x\ge-1\)

\(7+12\sqrt{x+1}=x+4\sqrt{x^2+3x+2}\)

\(\Leftrightarrow4\sqrt{\left(x+1\right)\left(x+2\right)}-12\sqrt{x+1}+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\sqrt{x+2}-3\right)+x-7=0\)

\(\Leftrightarrow4\sqrt{x+1}\left(\dfrac{x-7}{\sqrt{x+2}+3}\right)+x-7=0\)

\(\Leftrightarrow\left(x-7\right)\left(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1\right)=0\)

\(\Leftrightarrow x-7=0\) (do \(\dfrac{4\sqrt{x+1}}{\sqrt{x+2}+3}+1>0;\forall x\ge-1\))

\(\Rightarrow x=7\)

b.

ĐKXĐ: \(x\ne-\dfrac{1}{3}\)

\(\Rightarrow3x^2+3x+2=\left(3x+1\right)\sqrt{x^2+x+2}\)

\(\Leftrightarrow x^2+x+2-\left(3x+1\right)\sqrt{x^2+x+2}+2x^2+2x=0\)

Đặt \(\sqrt{x^2+x+2}=t\)

\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)

\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1+x-1}{2}=2x\\t=\dfrac{3x+1-\left(x-1\right)}{2}=x+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2x\left(x\ge0\right)\\\sqrt{x^2+x+2}=x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4x^2\left(x\ge0\right)\\x^2+x+2=x^2+2x+1\left(x\ge-1\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{3}\\\end{matrix}\right.\)

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

1.

6x + 1 ≥0

<=>6x≥-1

<=>x≥-1/6

2.

3x - 5 > 0 

<=> 3x > 5

<=> x > 5/3

3.

x - 7 > 0

<=> x > 7

4. 

-3x ≥0

<=>x≤0

a/ ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow\sqrt{x+9}-3+\sqrt{2x+4}-2=0\)

\(\Leftrightarrow\frac{x}{\sqrt{x+9}+3}+\frac{2x}{\sqrt{2x+4}+2}=0\)

\(\Leftrightarrow x\left(\frac{1}{\sqrt{x+9}+3}+\frac{2}{\sqrt{2x+4}+2}\right)=0\)

\(\Leftrightarrow x=0\)

b/ ĐKXĐ: \(x\ge\frac{7}{5}\)

\(\Leftrightarrow\sqrt{5x-7}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{3x+8}\)

\(\Leftrightarrow\frac{2x-12}{\sqrt{5x-7}+\sqrt{3x+5}}=\frac{2x-12}{\sqrt{5x-4}+\sqrt{3x+8}}\)

\(\Leftrightarrow2x-12=0\Rightarrow x=6\)

Bình phương đi bạn

TL:

1đk:x<1

.\(1+3x-1=9x^2\) 

     \(3x=9x^2\) 

   x=3x\(^2\) 

 =>x=0(ktm)  hoặc  x= \(\frac{1}{3}\left(tm\right)\) 

vậy x=\(\frac{1}{3}\) 

hc tốt:)