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)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

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

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

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

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)

Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)

2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)

Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)

\(\Rightarrow\left(x-2\right)^2=0\)

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

Vậy pt có no x=2

a)\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

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

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

Đặt \(x-3=t\) pt thành

\(\sqrt{t\left(t-6\right)}-t=0\)

\(\Leftrightarrow t^2-6t=t^2\)

\(\Leftrightarrow t=0\)\(\Rightarrow x-3=0\Leftrightarrow x=3\)

b)\(\sqrt{x^2-4}-x^2+4=0\)

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

Đặt \(\sqrt{x^2-4}=t\) pt thành

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

\(\Rightarrow\left[\begin{array}{nghiempt}t=1\\t=0\end{array}\right.\).

Với \(t=0\Rightarrow\sqrt{x^2-4}=0\Rightarrow x=\pm2\) 

Với \(t=1\Rightarrow\sqrt{x^2-4}=1\)\(\Rightarrow x=\pm\sqrt{5}\)

bạn đăg từng bài thui

mong mọi người giải giúp em vs 

6) \(\sqrt{x^2-4x+1}=x\left(x\ge0\right)\) 

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

\(\Leftrightarrow x^2-x^2=4x-1\)

\(\Leftrightarrow4x=1\)

\(\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\) 

8) \(\sqrt{x^2-x-6}=\sqrt{x-3}\left(x\ge3\right)\) 

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

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

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

9) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\left(x\ge1\right)\)

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

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=1+1\)

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