a:

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

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=0 hoặc x=6

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

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

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

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

=>x-1=1

=>x=2

c:

ĐKXĐ: x>4/5

PT \(\Leftrightarrow\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

d: ĐKXĐ: x-4>=0 và x+1>=0

=>x>=4

PT =>\(\left(\sqrt{x-4}+\sqrt{x+1}\right)^2=5^2=25\)

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

=>\(\sqrt{4\left(x^2-3x-4\right)}=25-2x+3=28-2x\)

=>\(\sqrt{x^2-3x-4}=14-x\)

=>x<=14 và x^2-3x-4=(14-x)^2=x^2-28x+196

=>x<=14 và -3x-4=-28x+196

=>x<=14 và 25x=200

=>x=8(nhận)

a) \(\sqrt{x^2-6x+9}=3\)

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

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

TH1: \(\left|x-3\right|=x-3\) với \(x\ge3\)

Pt trở thành:

\(x-3=3\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=3+3\)

\(\Leftrightarrow x=6\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x< 3\)

Pt trở thành:

\(-\left(x-3\right)=3\) (ĐK: \(x< 3\))

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=-3+3\)

\(\Leftrightarrow x=0\left(tm\right)\)

b) \(\sqrt{x+2\sqrt{x-1}}=2\) (ĐK: \(x\ge1\))

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

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

\(\Leftrightarrow4\left(x-1\right)=16-8x+x^2\)

\(\Leftrightarrow4x-4=16-8x+x^2\)

\(\Leftrightarrow x^2-12x+20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

c) \(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\) (ĐK: \(x\ge\dfrac{4}{5}\))

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

\(\Leftrightarrow5x-4=4x+8\)

\(\Leftrightarrow x=12\left(tm\right)\)

2.

\(DK:\hept{\begin{cases}x\ge-\frac{1}{5}\\x\ne0\end{cases}}\)

PT

\(\Leftrightarrow6+3\sqrt{5x+1}\left(\sqrt{5x+1}-1\right)=14\left(\sqrt{5x+1}-1\right)\)

\(\Leftrightarrow15x+23-17\sqrt{5x+1}=0\)

\(\Leftrightarrow\left(68-17\sqrt{5x+1}\right)+\left(15x-45\right)=0\)

\(\Leftrightarrow\frac{17\left(x-3\right)}{4+\sqrt{5x+1}}+15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{17}{4+\sqrt{5x+1}}+15\right)=0\)

Vi \(\frac{17}{4+\sqrt{5x+1}}+15>0\)

\(\Rightarrow x=3\left(n\right)\)

Vay nghiem cua PT la \(x=3\)

a/ ĐKXĐ: ...

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

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)\)

\(\Leftrightarrow6x^2+15x-26=0\)

b/ ĐKXĐ: ...

Đặt \(\sqrt[5]{\frac{16x}{x-1}}=a\)

\(a+\frac{1}{a}=\frac{5}{2}\Leftrightarrow a^2-\frac{5}{2}a+1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[5]{\frac{16x}{x-1}}=2\\\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}16x=32\left(x-1\right)\\16x=\frac{1}{32}\left(x-1\right)\end{matrix}\right.\)

c/ĐKXĐ: ...

\(\Leftrightarrow x^2-2x-\sqrt{6x^2-12x+7}=0\)

Đặt \(\sqrt{6x^2-12x+7}=a\ge0\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

\(\frac{a^2-7}{6}-a=0\Leftrightarrow a^2-6a-7=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=7\end{matrix}\right.\) \(\Rightarrow\sqrt{6x^2-12x+7}=7\)

\(\Leftrightarrow6x^2-12x-42=0\)

d/ \(\Leftrightarrow x^2+x+4-\sqrt{x^2+x+4}-2=0\)

Đặt \(\sqrt{x^2+x+4}=a>0\)

\(a^2-a-2=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\)

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

e/ \(\Leftrightarrow x^2+2x+\sqrt{3x^2+6x+4}-2=0\)

Đặt \(\sqrt{3x^2+6x+4}=a>0\Rightarrow x^2+2x=\frac{a^2-4}{3}\)

\(\frac{a^2-4}{3}+a-2=0\)

\(\Leftrightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{3x^2+6x+4}=2\Rightarrow3x^2+6x=0\)

ĐKXĐ:...

a/ \(\sqrt{2x^2+5x+2}=1+2\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow2x^2+5x+2=4\left(2x^2+5x-6\right)+1+4\sqrt{2x^2+5x-6}\)

\(\Leftrightarrow3\left(2x^2+6x-6\right)+4\sqrt{2x^2+5x-6}-7=0\)

Đặt \(\sqrt{2x^2+5x-6}=a\ge0\)

\(3a^2+4a-7=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{7}{3}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x^2+5x-6}=1\)

\(\Leftrightarrow2x^2+5x-7=0\)

a/ ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)

\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)

\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)

b/ ĐKXĐ: ...

\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)

Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)

\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)

\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)

c/ ĐKXĐ: ...

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

Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)

\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)

\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)

\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)

e/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x}{x-1}}=a>0\)

\(a+\frac{1}{a}=\frac{3}{\sqrt{2}}\Leftrightarrow a^2-\frac{3}{\sqrt{2}}a+1=0\)

\(\Rightarrow\left[{}\begin{matrix}a=\sqrt{2}\\a=\frac{\sqrt{2}}{2}\end{matrix}\right.\)

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

f/ ĐKXĐ: ...

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

Bình phương 2 vế:

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

\(\Rightarrow x=1\)

Lời giải:

1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$

PT $\Leftrightarrow x^2+5x+1=x+1$

$\Leftrightarrow x^2+4x=0$

$\Leftrightarrow x(x+4)=0$

$\Rightarrow x=0$ hoặc $x=-4$

Kết hợp đkxđ suy ra $x=0$

2. ĐKXĐ: $x\leq 2$

PT $\Leftrightarrow x^2+2x+4=2-x$

$\Leftrightarrow x^2+3x+2=0$

$\Leftrightarrow (x+1)(x+2)=0$

$\Leftrightarrow x+1=0$ hoặc $x+2=0$

$\Leftrightarrow x=-1$ hoặc $x=-2$
3.

ĐKXĐ: $-2\leq x\leq 2$

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

$\Leftrightarrow 2x+4=2-x$

$\Leftrightarrow 3x=-2$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

a/ Dặt \(\sqrt{x+1}=a\ge0\)

\(\Rightarrow4\sqrt{x+1}=x^2+5x+4\)

\(\Leftrightarrow4\sqrt{x+1}=\left(x+1\right)^2+3\left(x+1\right)\)

\(\Leftrightarrow4a=a^4+3a^2\)

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

\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+1}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)

b/ Đặt \(\hept{\begin{cases}\sqrt{4x+1}=a\ge0\\\sqrt{3x-2}=b\ge0\end{cases}}\)

\(\Rightarrow a^2-b^2=x+3\)

Từ đây ta có:

\(a-b=\frac{a^2-b^2}{5}\)

\(\Leftrightarrow\left(a-b\right)\left(5-a-b\right)=0\)

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

Thế vô làm tiếp

c/

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

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

\(\Leftrightarrow4x^4-20x^3+45x^2-55x+26=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(4x^2-8x+13\right)=0\)

Làm nốt

a.

ĐKXĐ: \(x\ge1\)

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

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

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

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

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

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)

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

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

\(\Leftrightarrow x=3\)

c.

ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)

\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)

\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)

Dấu "=" xảy ra khi và chỉ khi \(x=-1\)

d.

ĐKXĐ: \(x>1\)

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

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x-1}=b>0\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2-1}{a}=\dfrac{1-b^2}{b}\)

\(\Leftrightarrow a-\dfrac{1}{a}=\dfrac{1}{b}-b\)

\(\Leftrightarrow a+b-\dfrac{a+b}{ab}=0\)

\(\Leftrightarrow\left(a+b\right)\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow1-\dfrac{1}{ab}=0\)

\(\Leftrightarrow ab=1\)

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

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

\(\Leftrightarrow x=\sqrt[3]{2}\)

Lời giải:
a/ ĐKXĐ: $x\geq 0; y\geq 1$

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

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

Vì $(\sqrt{x}-1)^2\geq 0; (\sqrt{y-1}-2)^2\geq 0$ với mọi $x,y$ thuộc đkxđ

Do đó để tổng của chúng bằng $0$ thì:

$\sqrt{x}-1=\sqrt{y-1}-2=0$

$\Leftrightarrow x=1; y=5$

b. ĐKXĐ: $x\geq 0; y\geq 1; z\geq 2$

PT $\Leftrightarrow 2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z$

$\Leftrightarrow (x-2\sqrt{x}+1)+[(y-1)-2\sqrt{y-1}+1]+[(z-2)-2\sqrt{z-2}+1]=0$

$\Leftrightarrow (\sqrt{x}-1)^2+(\sqrt{y-1}-1)^2+(\sqrt{z-2}-1)^2=0$

$\Rightarrow \sqrt{x}-1=\sqrt{y-1}-1=\sqrt{z-2}-1=0$

$\Leftrightarrow x=1; y=2; z=3$