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)

\(\dfrac{1}{3}\sqrt[]{9x+9}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4x+4}{25}}=11\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt[]{9\left(x+1\right)}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4\left(x+1\right)}{25}}=11\)

\(\Leftrightarrow\sqrt[]{x+1}-2\sqrt[]{x+1}+\dfrac{16}{5}\sqrt[]{x+1}=11\)

\(\Leftrightarrow\dfrac{11}{5}\sqrt[]{x+1}=11\)

\(\Leftrightarrow\sqrt[]{x+1}=5\)

\(\Leftrightarrow x+1=25\)

\(\Leftrightarrow x=24\)

\(\sqrt{4x^2-4x+1}=3-x\left(x\in R\right)\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3-x\\ \Leftrightarrow2x-1=3-x\\ \Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\\ \sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\right)\left(x\ge-1\right)\\ \Leftrightarrow\sqrt{x+1}\left(\sqrt{9}+1+\sqrt{4}\right)=2\left(x+1\right)\\ \Leftrightarrow6\sqrt{x+1}=2\left(x+1\right)\\ \Leftrightarrow3\sqrt{x+1}=x+1\\ \Leftrightarrow\sqrt{x+1}\left(3-\sqrt{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{x+1}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x+1=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=8\left(tm\right)\end{matrix}\right.\)

a, ĐK: \(x\in R\)

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

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

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

TH1: \(\left\{{}\begin{matrix}2x-1\ge0\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{4}{3}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\1-2x=3-x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x=-2\end{matrix}\right.\Leftrightarrow x=-2\)

b, ĐK: \(x\ge-1\)

\(\sqrt{9x+9}+\sqrt{x+1}-\sqrt{4x+4}=2\left(x+1\text{​​}\right)\)

\(\Leftrightarrow3\sqrt{x+1}+\sqrt{x+1}-2\sqrt{x+1}=2\left(x+1\right)\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)

\(a,\sqrt{x+1}=\sqrt{2-x}\)

\(\Rightarrow x+1=2-x\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

a) \(ĐKXĐ:-1\le x\le2\)

Bình phương 2 vế ta có: 

\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )

Vậy \(x=\frac{1}{2}\)

b) \(ĐKXĐ:x\ge1\)

\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)

\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)

\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )

Vậy \(x=65\)

c) \(ĐKXĐ:x\ge1\)

\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)

\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)

\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)

\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )

Vậy \(x=5\)

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:

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

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: \(x\ge-1\))

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

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

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

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

\(\Leftrightarrow x+1=16\)

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

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

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

\(\Leftrightarrow x=1\)

Đăng 1 lúc mà nhiều thế. Lần sau đăng 1 câu thôi b.

b/ \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2+1}+\sqrt{\left(x-2\right)^2+4}+\sqrt{\left(x-2\right)^2+5}=3+\sqrt{5}\)

Ta có: \(VT\ge1+2+\sqrt{5}=3+\sqrt{5}\)

Dấu = xảy ra khi \(x=2\)

c/ \(\sqrt{2-x^2+2x}+\sqrt{-x^2-6x-8}=\sqrt{3-\left(x-1\right)^2}+\sqrt{1-\left(x+3\right)^2}\)

\(\le1+\sqrt{3}\)

Dấu = không xảy ra nên pt vô nghiệm

Câu d làm tương tự

\(a,\sqrt{x^2-4}-x^2+4=0\) 

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

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

\(\Leftrightarrow x^2-4-x^4+8x^2-16=0\)  

\(\Leftrightarrow-x^4-7x^2-20=0\) 

\(\Leftrightarrow-\left(x^4+7x^2+\frac{49}{4}\right)-\frac{31}{4}=0\) 

\(\Leftrightarrow-\left(x^2+\frac{7}{2}\right)^2=\frac{31}{4}\) 

\(\Leftrightarrow\left(x^2+\frac{7}{2}\right)=-\frac{31}{4}\) 

\(\Rightarrow\)pt vô nghiệm

a/ĐK: \(x\ge2\)

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

Đặt \(x^2-4=t\Rightarrow x^2=t+4\)

Thay vào,phương trình đã cho tương đương với: 

\(t=\sqrt{t}\Leftrightarrow t^2=t\Rightarrow\orbr{\begin{cases}t=1\\t=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-4=1\\x^2-4=0\end{cases}}\)

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

a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

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

b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

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

c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)

TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)

TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)

Vậy x=0,5...

d, đk \(x\ge-1\)

=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)

\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)

a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b) Ta có: \(\sqrt{4x^2-4x+1}=3\)

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

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

c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)

\(\Leftrightarrow\left|x-3\right|=4-3x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)