Rút gọn biểu thức chứa căn bậc hai

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

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

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

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

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

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

b) Ta có: \(x=3-2\sqrt{2}\)

\(=2-2\cdot\sqrt{2}\cdot1+1\)

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

Thay \(x=\left(\sqrt{2}-1\right)^2\) vào biểu thức \(P=\dfrac{\sqrt{2}-\sqrt{x}}{\sqrt{x}}\), ta được: 

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

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

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

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

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

Vậy: Khi \(x=3-2\sqrt{2}\) thì \(P=\sqrt{2}+1\)

Lời giải:

PT $\Leftrightarrow (4x^2+y^2-4xy)+9y^2+12x+6y+13=0$

$\Leftrightarrow (2x-y)^2+6(2x-y)+9y^2+12y+13=0$

$\Leftrightarrow (2x-y)^2+6(2x-y)+9+(9y^2+12y+4)=0$

$\Leftrightarrow (2x-y+3)^2+(3y+2)^2=0$

$\Rightarrow (2x-y+3)^2=(3y+2)^2=0$

$\Rightarrow y=-\frac{2}{3}; x=\frac{-11}{6}$

Ta có: \(P=\dfrac{4\sqrt{x}+3}{x+\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

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

Để P nguyên thì \(\sqrt{x}+3⋮\sqrt{x}\)

mà \(\sqrt{x}⋮\sqrt{x}\)

nên \(3⋮\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}\inƯ\left(3\right)\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;-1;3;-3\right\}\)

mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\in\left\{1;3\right\}\)

\(\Leftrightarrow x\in\left\{1;9\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{1;9\right\}\)

Vậy: Để P nguyên thì \(x\in\left\{1;9\right\}\)

Ta có: \(\sqrt{5-2\sqrt{6}}+2\sqrt{2}\)

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

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

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

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

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

Bạn có thể vt rõ hơn được ko?

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

     \(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)

     \(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}-1}{a-1}\)

     \(=\dfrac{1}{\sqrt{a}}\)

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

     \(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)

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

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

mai mk thi rùi cầu cho các bạn trai xinh gái đẹp giúp mk với huhu

a) Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\), ta được: 

\(A=\left(\sqrt{\dfrac{1}{4}}+1\right):\left(\sqrt{\dfrac{1}{4}}-2\right)\)

\(\Leftrightarrow A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)\)

\(\Leftrightarrow A=\dfrac{3}{2}:\dfrac{-3}{2}\)

\(\Leftrightarrow A=\dfrac{3}{2}\cdot\dfrac{2}{-3}=\dfrac{3}{-3}=-1\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì A=-1

b) Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+4\sqrt{x}-3\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

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

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

Lời giải:a) Với \(x=\frac{1}{4}\Rightarrow \sqrt{x}=\frac{1}{2}\)

Khi đó: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=-1\)

b) \(B=\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{\sqrt{x}-8}{(\sqrt{x}-2)(\sqrt{x}-3)}=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-3)(\sqrt{x}-2)}+\frac{\sqrt{x}-8}{(\sqrt{x}-2)(\sqrt{x}-3)}\)

\(=\frac{x-4+\sqrt{x}-8}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{x+\sqrt{x}-12}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{(\sqrt{x}-3)(\sqrt{x}+4)}{(\sqrt{x}-3)(\sqrt{x}-2)}=\frac{\sqrt{x}+4}{\sqrt{x}-2}\)