\(\Leftrightarrow\left|2x+1\right|=\left|x+6\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+6\\2x+1=-x-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{3}\end{matrix}\right.\)

ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+4x+1}=\sqrt{x^2+12x+36}\\ \Leftrightarrow\left|2x+1\right|=\left|x+6\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=x+6\\2x+1=-x-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{3}\end{matrix}\right.\)

d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

`a, <=> 5/3 . 3sqrt(x^2+2) + 3/2.2sqrt(x^2+2)-7sqrt6=sqrt(x^2+2)`

`= (5+3-1)sqrt(x^2+2)=7sqrt6`

`<=> 7sqrt(x^2+2)=7sqrt6`.

`<=> x^2+2=36`.

`<=> x^2=34`.

`<=> x=+-sqrt(34)`.

Vậy...

`b, sqrt(4x^2-12x+9)-6=0`

`<=> |2x-3|=6`.

`@ x >=3/2 <=> 2x-3=6.`

`<=> x=9/2 (tm)`.

`@x <3/2 <=> 3-2x=6`

`<=> 2x=-3`

`<=> x=-3/2.`

Vậy...

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{4}$

PT $\Leftrightarrow \sqrt{3x+4}=\sqrt{12x+7}+\sqrt{4x+1}$

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

$\Leftrightarrow 13x+4+2\sqrt{(12x+7)(4x+1)}=0$

Với $x\geq \frac{-1}{4}$ dễ thấy vế trái $>0$. Do đó pt đã cho vô nghiệm.

a. ĐKXĐ: $x\geq 2$ hoặc $x=1$

PT $\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}$

$\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=3\end{matrix}\right.\) (đều thỏa mãn)

b.

PT $\Leftrightarrow \sqrt{(x-2)^2}=\sqrt{(2x-3)^2}$

$\Leftrightarrow |x-2|=|2x-3|$

\(\Leftrightarrow \left[\begin{matrix} x-2=2x-3\\ x-2=3-2x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\)

c. ĐKXĐ: $x=2$ hoặc $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}(\sqrt{x-3}-1)=0$

\(\Leftrightarrow \left[\begin{matrix} \sqrt{x-2}=0\\ \sqrt{x-3}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=4\end{matrix}\right.\) (đều tm)

d.

PT $\Leftrightarrow \sqrt{(2x-1)^2}=\sqrt{(x-3)^2}$

$\Leftrightarrow |2x-1|=|x-3|$

\(\Leftrightarrow \left[\begin{matrix} 2x-1=x-3\\ 2x-1=3-x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\)

a: Ta có: \(\sqrt{x^2-3x+2}=\sqrt{x-1}\)

\(\Leftrightarrow x^2-3x+2=x-1\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=-x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{x^2-5x+6}=\sqrt{x-2}\)

\(\Leftrightarrow x^2-5x+6=x-2\)

\(\Leftrightarrow x^2-6x+8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

b) \(1+4x-3|x+2|+4=0\)

\(\Leftrightarrow4x-3|x+2|=-5\left(1\right)\)

TH1: Với \(|x+2|=x+2\)thay vào (1) ta được:

\(4x-3\left(x+2\right)=-5\)

\(\Leftrightarrow4x-3x-6=-5\)

\(\Leftrightarrow x=1\)(chọn tự thử lại nhé nó =0 )

TH2: Với \(|x+2|=-x-2\)thay vào (1) ta được: 

\(4x-3\left(-x-2\right)=-5\)

\(\Leftrightarrow4x+3x+6=-5\)

\(\Leftrightarrow7x=-11\)

\(\Leftrightarrow x=\frac{-11}{7}\)( loại tự thử lại nhé nó ko =0 )

Vậy x=1

`a)\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=12`     `ĐK: x >= 0`

`<=>\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12`

`<=>12\sqrt{3x}=12`

`<=>\sqrt{3x}=1`

`<=>3x=1<=>x=1/3` (t/m)

`b)5\sqrt{9x+9}-2\sqrt{4x+4}+\sqrt{x+1}=36`   `ĐK: x >= -1`

`<=>15\sqrt{x+1}-4\sqrt{x+1}+\sqrt{x+1}=36`

`<=>12\sqrt{x+1}=36`

`<=>\sqrt{x+1}=3`

`<=>x+1=9`

`<=>x=8` (t/m)