\(\sqrt{4x^2-4x+1}=\sqrt{x^2+10x+25}\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-\left(x+5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=x+5\\2x-1=-x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)

a) 

\(\sqrt{x+3}+2\sqrt{4\left(x+3\right)}-\frac{1}{3}\sqrt{9\left(x+3\right)}=8\)  

\(\sqrt{x+3}+2\cdot2\sqrt{x+3}-\frac{1}{3}\cdot3\sqrt{x+3}=8\)    

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

\(4\sqrt{x+3}=8\)          

\(\sqrt{x+3}=2\) 

\(\orbr{\begin{cases}2\ge0\left(llđ\right)\\x+3=2^2\end{cases}}\) 

\(x+3=4\) 

\(x=1\) 

b) 

\(\orbr{\begin{cases}x^2+10x+25\ge0\\4x^2-4x+1=x^2+10x+25\end{cases}}\) 

\(\orbr{\begin{cases}\left(x+5\right)^2\ge0\left(lld\right)\\3x^2-6x-24=0\end{cases}}\) 

\(\orbr{\begin{cases}x=6\\x=-\frac{4}{3}\end{cases}}\)        

1/ ĐKXĐ: $4x^2-4x-11\geq 0$

PT $\Leftrightarrow \sqrt{4x^2-4x-11}=2(4x^2-4x-11)-6$

$\Leftrightarrow a=2a^2-6$ (đặt $\sqrt{4x^2-4x-11}=a, a\geq 0$)

$\Leftrightarrow 2a^2-a-6=0$

$\Leftrightarrow (a-2)(2a+3)=0$

Vì $a\geq 0$ nên $a=2$

$\Leftrightarrow \sqrt{4x^2-4x-11}=2$

$\Leftrightarrow 4x^2-4x-11=4$

$\Leftrightarrow 4x^2-4x-15=0$
$\Leftrightarrow (2x-5)(2x+3)=0$

$\Rightarrow x=\frac{5}{2}$ hoặc $x=\frac{-3}{2}$ (tm)

2/ ĐKXĐ: $x\in\mathbb{R}$

PT $\Leftrightarrow \sqrt{3x^2+9x+8}=\frac{1}{3}(3x^2+9x+8)-\frac{14}{3}$

$\Leftrightarrow a=\frac{1}{3}a^2-\frac{14}{3}$ (đặt $\sqrt{3x^2+9x+8}=a, a\geq 0$)

$\Leftrightarrow a^2-3a-14=0$

$\Rightarrow a=\frac{3+\sqrt{65}}{2}$ (do $a\geq 0$)

$\Leftrightarrow 3x^2+9x+8=\frac{37+3\sqrt{65}}{2}$

$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{23+2\sqrt{65}})$

3. ĐKXĐ: $x^2+3x\geq 0$

PT $\Leftrightarrow 10-(x^2+3x)=\sqrt{x^2+3x}$

$\Leftrightarrow 10-a^2=a$ (đặt $\sqrt{x^2+3x}=a, a\geq 0$)

$\Leftrightarrow a^2+a-10=0$

$\Rightarrow a=\frac{-1+\sqrt{41}}{2}$

$\Leftrightarrow x^2+3x=a^2=\frac{21-\sqrt{41}}{2}$

$\Rightarrow x=\frac{1}{2}(-3\pm \sqrt{51-2\sqrt{41}})$ (đều tm)

\(2\sqrt{9\left(x-3\right)}-\sqrt{4\left(x-3\right)}=10+\frac{1}{2}\)

\(6\sqrt{\left(x-3\right)}-2\sqrt{\left(x-3\right)}=\frac{21}{2}\)

\(4\sqrt{\left(x-3\right)}=\frac{21}{2}\)

\(\sqrt{\left(x-3\right)}=\frac{21}{8}\)

\(x-3=\frac{441}{64}\)

\(x=\frac{633}{64}\)

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

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

\(\Leftrightarrow x=1\)

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

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

\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

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)

đề bài như trên

\(\Leftrightarrow\sqrt{9\left(x-3\right)}+\sqrt{x-3}-\frac{1}{2}\sqrt{4\left(x-3\right)}=7\)

\(\Leftrightarrow3\sqrt{x-3}+\sqrt{x-3}-\frac{1}{2}.2\sqrt{x-3}=7\)

\(\Leftrightarrow3\sqrt{x-3}=7\)

\(\Leftrightarrow\sqrt{x-3}=\frac{7}{3}\left(đk:x\ge3\right)\)

\(\Leftrightarrow x-3=\frac{49}{9}=>x=\frac{76}{9}\left(thoảman\right)\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

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

\(ĐK:x\ge5\)

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

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

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

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

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

cho mình hỏi hai ý đầu thôi, hai ý sau mình giải ra rồi. Thanks Zero ~