a.

$x^2-11=0$

$\Leftrightarrow x^2=11$

$\Leftrightarrow x=\pm \sqrt{11}$

b. $x^2-12x+52=0$

$\Leftrightarrow (x^2-12x+36)+16=0$

$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)

Vậy pt vô nghiệm.

c.

$x^2-3x-28=0$

$\Leftrightarrow x^2+4x-7x-28=0$

$\Leftrightarrow x(x+4)-7(x+4)=0$

$\Leftrightarrow (x+4)(x-7)=0$

$\Leftrightarrow x+4=0$ hoặc $x-7=0$

$\Leftrightarrow x=-4$ hoặc $x=7$

d.

$x^2-11x+38=0$

$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$

$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)

Vậy pt vô nghiệm

e.

$6x^2+71x+175=0$

$\Leftrightarrow 6x^2+21x+50x+175=0$

$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$

$\Leftrightarrow (3x+25)(2x+7)=0$

$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$

$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$

f.

$x^2-(\sqrt{2}+\sqrt{8})x+4=0$

$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$

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

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

$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$

$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$

g.

$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$

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

$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$

$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$

$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$

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

a) \(A=\dfrac{x\sqrt{y}+y\sqrt{x}}{x+2\sqrt{xy}+y}\)

\(A=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

\(A=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

b) \(B=\dfrac{x\sqrt{y}-y\sqrt{x}}{x-2\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

c) \(C=\dfrac{3\sqrt{a}-2a-1}{4a-4\sqrt{a}+1}\)

\(C=\dfrac{-\left(2a-3\sqrt{a}+1\right)}{\left(2\sqrt{a}\right)^2-2\sqrt{a}\cdot2\cdot1+1^2}\)

\(C=\dfrac{-\left(\sqrt{a}-1\right)\left(2\sqrt{a}-1\right)}{\left(2\sqrt{a}-1\right)^2}\)

\(C=\dfrac{-\sqrt{a}+1}{2\sqrt{a}-1}\)

d) \(D=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)

\(D=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{\sqrt{a}-2}\)

\(D=\sqrt{a}+2-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)

\(D=\left(\sqrt{a}+2\right)-\left(\sqrt{a}+2\right)\)

\(D=0\)

a) \(P=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

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

\(P=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)\(P=\left(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)

\(P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

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

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

\(P=\dfrac{2}{x+\sqrt{x}+1}\)

b) Mà với \(x\ge0\) và \(x\ne1\) thì 

\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)

a: \(P=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

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

b: x+căn x+1+1>=1>0

2>0

=>P>0 với mọi x thỏa mãn x>=0 và x<>1

a: ĐKXĐ: x>=1

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

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

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

=>\(\dfrac{3}{2}\sqrt{x-1}=3\)

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

=>x-1=4

=>x=5(nhận)

b: \(\sqrt{x^2-4x+4}+x-2=0\)

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

=>|x-2|=-(x-2)

=>x-2<=0

=>x<=2

c: 

ĐKXĐ: 7-x>=0

=>x<=7

\(\sqrt{7-x}+1=x\)

=>\(\sqrt{7-x}=x-1\)

=>\(\left\{{}\begin{matrix}x-1>=0\\7-x=x^2-2x+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1< =x< =7\\x^2-2x+1-7+x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1< =x< =7\\x^2-x-6=0\end{matrix}\right.\Leftrightarrow x=3\)

a) Ta có: \(\sqrt{49\left(x^2-2x+1\right)}-35=0\)

\(\Leftrightarrow7\left|x-1\right|=35\)

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

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

b)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

Ta có: \(\sqrt{x^2-9}-5\sqrt{x+3}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{x-3}=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=28\left(nhận\right)\end{matrix}\right.\)

c) ĐKXĐ: \(x\ge0\)

Ta có: \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)

\(\Leftrightarrow x-1=x+\sqrt{x}-6\)

\(\Leftrightarrow\sqrt{x}-6=-1\)

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

hay x=25(nhận)

a: \(x^2-2-x+\sqrt{2}=0\)

=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)

=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-1\right)=0\)

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

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

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

\(=12+8\sqrt{2}+4\left(\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)\)

\(=12+8\sqrt{2}+4\left(6+\sqrt{2}-3\sqrt{2}-1\right)\)

\(=12+8\sqrt{2}+24-8\sqrt{2}-4=32>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2\left(1+\sqrt{2}\right)-4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=1\\x_2=\dfrac{2\left(1+\sqrt{2}\right)+4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=-7-4\sqrt{2}\end{matrix}\right.\)

1. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$

$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$

$\Leftrightarrow 22=10\sqrt{x-4}$

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

$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$

(thỏa mãn)

2. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$

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

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

$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$

$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)

3. ĐKXĐ: $x\geq 3$

Bình phương 2 vế thu được:

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

$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$

$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$

$\Leftrightarrow (x-4)(7x+4)=0$

Do $x\geq 3$ nên $x=4$

Thử lại thấy thỏa mãn

Vậy $x=4$

4. ĐKXĐ: $x\geq 4$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$

$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$

Ta thấy, với mọi $x\geq 4$ thì:

$(\sqrt{x}-2)^2\ge 0$

$2021\sqrt{x-4}\geq 0$ 

Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$

$\Leftrightarrow x=4$ (tm)

\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)

a: \(\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\)

\(=\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+35}{x-25}\)

b: \(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\)

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

c: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)

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

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

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

d: \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

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

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

=>x-2=0 hoặc x+2=1/9

=>x=-17/9(loại) hoặc x=2

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

=>x^2-1=0 hoặc x^2-1=1

=>x^2=1 hoặc x^2=2

=>\(x\in\left\{1;-1;\sqrt{2};-\sqrt{2}\right\}\)