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

Vì \(VT\ge2\Rightarrow VP\ge2\Rightarrow x\ge\dfrac{2}{3}\)

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

\(\Rightarrow2x-1+2=3x\left(x\ge\dfrac{2}{3}\right)\Rightarrow x=1\)

\(7\sqrt{a}-5b\sqrt{16a^3}+4a\sqrt{25ab^2}-3\sqrt{16a}\)

\(=7\sqrt{a}-20ab\sqrt{a}+20ab\sqrt{a}-12\sqrt{a}=-5\sqrt{a}\)

\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)

\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)

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

\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)

\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)

a: 

Sửa đề: a+2căn a+8

\(=\dfrac{5a+10\sqrt{a}-3\sqrt{a}-6+3a-6\sqrt{a}-a-2\sqrt{a}-8}{\left(a-4\right)}\)

\(=\dfrac{7a-\sqrt{a}-14}{\left(a-4\right)}\)

b: A>0

=>(7a-căn a-14)/(a-4)>0

=>a>4 hoặc 0<a<(1+căn 393)/14

\(a,=\left|2-\sqrt{3}\right|=2-\sqrt{3}\\ b,=\left|3-\sqrt{11}\right|=\sqrt{11}-3\\ c,=2\left|a\right|=2a\\ d,=3\left|a-2\right|=3\left(2-a\right)\left(a< 0\Leftrightarrow a-2< 0\right)\)

a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)

b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)

c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)

Với `a > 0,b >= 0` có:

`Bth=[a\sqrt{b}+b]/[a-b] . \sqrt{[b(a+b-2\sqrt{ab})]/[a^2+2a\sqrt{b}+b]} . (\sqrt{a}+\sqrt{b})`

 `=[\sqrt{b}(a+\sqrt{b})]/[a-b].\sqrt{[b(\sqrt{a}-\sqrt{b})^2]/[(a+\sqrt{b})^2]}.(\sqrt{a}+\sqrt{b})`

`=[\sqrt{b}(a+\sqrt{b})|\sqrt{a}-\sqrt{b}|.\sqrt{b}.(\sqrt{a}+\sqrt{b})]/[(a-b)(a+\sqrt{b})]`

`=[b|\sqrt{a}-\sqrt{b}|]/[\sqrt{a}-\sqrt{b}]`

`={(b\text{ nếu }\sqrt{a} >= \sqrt{b}),(-b\text{ nếu }\sqrt{a} < \sqrt{b}):}`

\(\dfrac{2\sqrt{a}}{\sqrt{a}+3}+\dfrac{\sqrt{a}+1}{\sqrt{a}-3}+\dfrac{3+7\sqrt{a}}{9-a}\)

\(=\dfrac{2\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}+\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}-\dfrac{3+7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

\(=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)

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

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

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

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

\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)

=2

Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)

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

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