a: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>2x=-2 hoặc 2x=4

=>x=2 hoặc x=-1

c: \(\Leftrightarrow\left|x-3\right|=11-x\)

=>x<=11 và (x-3)^2=(11-x)^2

=>x<=11 và x^2-6x+9=x^2-22x+121

=>x<=11 và 16x=112

=>x=7

d:

ĐKXĐ: 3x+19>=0

=>x>=-19/3

PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9

=>x>=-3 và x^2+6x+9-3x-19=0 

=>x>=-3 và (x+5)(x-2)=0

=>x=2

e: =>\(\sqrt{x^2+x+5}=x+1\)

=>x>=-1 và x^2+x+5=x^2+2x+1

=>x>=-1 và 2x+1=x+5

=>x=4

1:

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

=>x-3=0 hoặc \(\sqrt{x+3}=2\)

=>x=3 hoặc x+3=4

=>x=1(loại) hoặc x=3(nhận)

2:

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

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

=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)

=>4(12x^2-16x+3x-4)=(7x-6)^2

=>49x^2-84x+36=48x^2-52x-16

=>-84x+36=-52x-16

=>-32x=-52

=>x=13/8

3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)

=>|x-5|=5-x

=>x-5<=0

=>x<=5

4: \(\Leftrightarrow\left|x-4\right|=x+2\)

=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)

=>x>=-2 và -8x+16=4x+4

=>x=1

a) \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(x\ge0;x\ne0\right)\)

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

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x-3}\right)}\)

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

b) \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\left(x\ge0;x\ne1\right)\)

\(=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

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

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

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

c) \(\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne1\right)\)

\(=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}+\dfrac{2.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

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

\(\dfrac{1}{\sqrt{x}+1}\)

\(A\sqrt{3}=\sqrt{3x+6\sqrt{3x-9}}+\sqrt{3x-6\sqrt{3x-9}}\)

\(=\sqrt{3x-9+6\sqrt{3x-9}+9}+\sqrt{3x-9-6\sqrt{3x-9}+9}\)

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

\(=\left|\sqrt{3x-9}+3\right|+\left|\sqrt{3x-9}-3\right|\)

Do \(x\ge6\Rightarrow\sqrt{3x-9}-3\ge0\)

\(\Rightarrow A\sqrt{3}=\sqrt{3x-9}+3+\sqrt{3x-9}-3=2\sqrt{3x-9}\ge6\)

\(\Rightarrow A\ge\frac{6}{\sqrt{3}}=2\sqrt{3}\)

Dấu "=" xảy ra khi \(x=6\)

a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)

\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)

\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)

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

Đến đây em xét các trường hợp rồi bình phương lên là được nha

a/ ĐKXĐ: $x\geq 1$

Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:

$a+b+2ab=6-(a^2+b^2)$

$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$

$\Leftrightarrow (a+b)^2+(a+b)-6=0$

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

Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$

$\Leftrightarrow a+b=2$

Mà $b^2-a^2=(x+3)-(x-1)=4$

$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$

$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$

$\Leftrightarrow x=1$ (tm)

b/

ĐKXĐ: $x\geq 1$

Đặt $\sqrt{3x-2}=a; \sqrt{x-1}=b(a,b\geq 0)$. Khi đó pt đã cho trở thành:

$a+b=a^2+b^2-6+2ab$

$\Leftrightarrow a^2+b^2+2ab-(a+b)-6=0$

$\Leftrightarrow (a+b)^2-(a+b)-6=0$

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

Hiển nhiên $a+b+2>0$ với mọi $a,b\geq 0$

Do đó $a+b-3=0\Leftrightarrow a+b=3$

$\Leftrightarrow b=3-a$.

Ta thấy $a^2-3b^2=1$. Thay $b=3-a$ vô thì:

$a^2-3(3-a)^2=1$

$\Leftrightarrow (a-2)(a-7)=0$

$\Leftrightarrow a=2$ hoặc $a=7$

Vì $a+b=3$ mà $a,b>0$ nên $a,b<3$. Do đó $a=2$

$\Leftrightarrow \sqrt{3x-2}=2$ 

$\Leftrightarrow x=2$ 

\(a,\sqrt{x^4}=7\Leftrightarrow x^2=7\Leftrightarrow x=\pm\sqrt{7}\)

\(Dk:x\ge\frac{2}{3};\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\left(tm\right)\)

\(dk:x\ge\frac{3}{2};\sqrt{2x-3}=\sqrt{x-1}\Leftrightarrow2x-3=x-1\Leftrightarrow x=2\left(tm\right)\)

\(dk:x\ge0;x-10\sqrt{x}+25=0\Leftrightarrow\left(\sqrt{x}-5\right)^2=0\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\left(tm\right)\)

\(\sqrt{2x}< 3\Leftrightarrow\sqrt{2}.\sqrt{x}< 3\Leftrightarrow0\le\sqrt{x}< \sqrt{4,5}\Leftrightarrow0\le x< 4,5\)

\(h,dk:x\ge3;\sqrt{\left(x-1\right)^2}=3x-9\Leftrightarrow\left|x-1\right|=3x-9\Leftrightarrow x-1=3x-9\left(x\ge3\right)\Leftrightarrow x=4\left(tm\right)\)

2:

ĐKXĐ: x>=3

 \(\Leftrightarrow\sqrt{x-3+2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot\sqrt{3}+3}=2\sqrt{3}\)

=>\(\left|\sqrt{x-3}+\sqrt{3}\right|+\left|\sqrt{x-3}-\sqrt{3}\right|=2\sqrt{3}\)

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

\(\Leftrightarrow\sqrt{x-3}+\left|\sqrt{x-3}-\sqrt{3}\right|=\sqrt{3}\)(1)

TH1: x>=6

(1) trở thành \(\sqrt{x-3}+\sqrt{x-3}-\sqrt{3}=\sqrt{3}\)

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

=>x-3=3

=>x=6(nhận)

TH2: 3<=x<6

Phương trình (1) sẽ là;

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

=>\(\sqrt{3}=\sqrt{3}\)(luôn đúng)

1:

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{8^2-\left(2\sqrt{10+2\sqrt{5}}\right)^2}\)

\(=16+2\cdot\sqrt{64-4\cdot\left(10+2\sqrt{5}\right)}\)

\(=16+2\cdot\sqrt{24-8\sqrt{5}}\)

\(=16+2\cdot\sqrt{20-2\cdot2\sqrt{5}\cdot2+4}\)

\(=16+2\cdot\sqrt{\left(2\sqrt{5}-2\right)^2}\)

\(=16+2\cdot\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=10+2\cdot\sqrt{10}\cdot\sqrt{2}+2\)

\(=\left(\sqrt{10}+\sqrt{2}\right)^2\)

=>\(A=\sqrt{10}+\sqrt{2}\)

a

\(\sqrt{9\left(2-3x\right)^2}=6\\ \Leftrightarrow3\left|2-3x\right|=6\\ \Leftrightarrow\left|2-3x\right|=2\)

Với \(x\le\dfrac{2}{3}\) thì PT trở thành:

\(2-3x=2\\ \Leftrightarrow3x=0\\ \Leftrightarrow x=0\left(nhận\right)\)

Với \(x>\dfrac{2}{3}\) thì PT trở thành:

\(3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\left(nhận\right)\)

b

ĐK: \(x\ge-\dfrac{3}{2}\)

\(\sqrt{4x^2-9}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{\left(2x\right)^2-3^2}=2\sqrt{2x+3}\\ \Leftrightarrow\sqrt{2x-3}.\sqrt{2x+3}-2\sqrt{2x+3}=0\\ \Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\2x-3=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\left(nhận\right)\\x=\dfrac{7}{2}\left(nhận\right)\end{matrix}\right.\)

c

ĐK: \(x\ge3\)

\(\sqrt{10\left(x-3\right)}=\sqrt{20}\\ \Leftrightarrow10\left(x-3\right)=20\\ \Leftrightarrow x-3=2\\ \Leftrightarrow x=5\left(nhận\right)\)

d

\(\sqrt{x^2+6x+9}=3x-6\\ \Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-6\\ \Leftrightarrow\left|x+3\right|=3x-6\)

Với \(x\ge-3\) thì PT trở thành:

\(x+3=3x-6\\ \Leftrightarrow x+3-3x+6=0\\ \Leftrightarrow-2x+9=0\\ \Leftrightarrow x=\dfrac{9}{2}\left(nhận\right)\)

Với \(x< -3\) thì PT trở thành:

\(-x-3=3x-6\\ \Leftrightarrow-x-3-3x+6=0\\ \Leftrightarrow-2x+3=0\\ \Leftrightarrow x=\dfrac{3}{2}\left(loại\right)\)