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a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)

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

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

b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)

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

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

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

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

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

22 tháng 6 2017

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

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

16 tháng 11 2021

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

31 tháng 5 2017

ĐKXĐ: \(x\ge0\)

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

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

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

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

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

2 tháng 8 2017

ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

Ta có \(A=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-\frac{x-\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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