a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

=>\(\dfrac{3}{2}\sqrt{x-3}=3\)

=>\(\sqrt{x-3}=2\)

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

c.

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

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

\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)

\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))

\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)

\(\Rightarrow x^3+7x^2+4x-24=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)

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

a.

\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)

Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)

Ta có:

\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)

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

\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

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

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

c: Ta có: \(\sqrt{x-1}+\sqrt{9x-9}-\sqrt{4x-4}=4\)

\(\Leftrightarrow2\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=4\)

hay x=5

e: Ta có: \(\sqrt{4x^2-28x+49}-5=0\)

\(\Leftrightarrow\left|2x-7\right|=5\)

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

a. ĐKXĐ: $x\in\mathbb{R}$

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

$\Leftrightarrow |x-2|=2-x$
$\Leftrightarrow 2-x\geq 0$

$\Leftrightarrow x\leq 2$

b. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{4}.\sqrt{x-2}-\frac{1}{5}\sqrt{25}.\sqrt{x-2}=3\sqrt{x-2}-1$

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

$\Leftrightarrow 1=2\sqrt{x-2}$

$\Leftrightarrow \frac{1}{2}=\sqrt{x-2}$

$\Leftrightarrow \frac{1}{4}=x-2$

$\Leftrightarrow x=\frac{9}{4}$ (tm)

c. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}+\sqrt{9}.\sqrt{x-1}-\sqrt{4}.\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}+3\sqrt{x-1}-2\sqrt{x-1}=4$

$\Leftrightarrow 2\sqrt{x-1}=4$

$\Leftrightarrow \sqrt{x-1}=2$

$\Leftrightarrow x-1=4$

$\Leftrightarrow x=5$ (tm)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-2}-4\sqrt{\frac{4}{9}}\sqrt{x-2}+\sqrt{9}.\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{1}{2}\sqrt{x-2}-\frac{8}{3}\sqrt{x-2}+3\sqrt{x-2}-5=0$

$\Leftrightarrow \frac{5}{6}\sqrt{x-2}-5=0$

$\Leftrightarrow \sqrt{x-2}=6$

$\Leftrightarrow x-2=36$

$\Leftrightarrow x=38$ (tm)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)

`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)

`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`

`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\) 

`b)sqrt{x-3}/sqrt{2x+1}=2`

ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)

`<=>x>=3`

`pt<=>sqrt{x-3}=2sqrt{2x+1}`

`<=>x-3=8x+4`

`<=>7x=7`

`<=>x=1(l)`

`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`

`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`

`<=>|x-1|+|x-2|=3`

`**x>=2`

`pt<=>x-1+x-2=3`

`<=>2x=6`

`<=>x=3(tm)`

`**x<=1`

`pt<=>1-x+2-x=3`

`<=>3-x=3`

`<=>x=0(tm)`

`**1<=x<=2`

`pt<=>x-1+2-x=3`

`<=>=-1=3` vô lý

Vậy `S={0,3}`

b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)

\(\Leftrightarrow x^2-6x+9=3\)

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

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)

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

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

\(\Leftrightarrow x=1\)

b: Ta có: \(\sqrt{4x+8}+\dfrac{1}{3}\sqrt{9x+18}=3\sqrt{\dfrac{x+2}{4}}+\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}+\dfrac{1}{3}\cdot3\sqrt{x+2}-\dfrac{3}{2}\sqrt{x+2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}\cdot\dfrac{3}{2}=\sqrt{2}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{2\sqrt{2}}{3}\)

\(\Leftrightarrow x+2=\dfrac{8}{9}\)

hay \(x=-\dfrac{10}{9}\)