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Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)
b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
a: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Bài 1:
\(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
\(=\left(\dfrac{x}{\left(x-7\right)\left(x+7\right)}-\dfrac{x-7}{x\cdot\left(x+7\right)}\right)\cdot\dfrac{x^2+7x}{2x-7}+\dfrac{x}{-\left(x-7\right)}\)
\(=\dfrac{x^2-\left(x-7\right)^2}{x\cdot\left(x-7\right)\left(x+7\right)}\cdot\dfrac{x\cdot\left(x+7\right)}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-\left(x-7\right)\right)\cdot\left(x+x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{\left(x-x+7\right)\cdot\left(2x-7\right)}{x-7}\cdot\dfrac{1}{2x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7}{x-7}-\dfrac{x}{x-7}\)
\(=\dfrac{7-x}{x-7}\)
\(=\dfrac{-\left(x-7\right)}{x-7}\)
\(=-1\)
A = \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x}{\left(x+7\right)\left(x-7\right)}-\dfrac{x-7}{x\left(x+7\right)}\right):\dfrac{2x-7}{x\left(x+7\right)}+\dfrac{x}{7-x}\)
A = \(\left(\dfrac{x^2-\left(x-7\right)^2}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{2x-7}{x\left(x+7\right)}-\dfrac{x}{x-7}\)
A = \(\left(\dfrac{x^2-\left(x^2-14x+49\right)}{\left(x+7\right)\left(x-7\right)x}\right):\dfrac{\left(2x-7\right)\left(x-7\right)-\left(x^3+7x^2\right)}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}:\dfrac{-x^3-5x^2-21x+49}{\left(x+7\right)\left(x-7\right)x}\)
A = \(\dfrac{14x-49}{\left(x+7\right)\left(x-7\right)x}.\dfrac{\left(x+7\right)\left(x-7\right)x}{-x^3-5x^2-21x+49}\)
A = \(\dfrac{14x-49}{-x^3-5x^2-21x+49}\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
\(a,VT=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{\left(x+1\right)\left(1-3x\right)}{3x}\right)\cdot\dfrac{x}{x-1}\\ =\left(\dfrac{2}{3x}-\dfrac{2-6x}{3x}\right)\cdot\dfrac{x}{x-1}=\dfrac{6x}{3x}\cdot\dfrac{x}{x-1}=\dfrac{2}{x-1}=VP\left(x\ne0;x\ne1\right)\)
\(b,VT=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}=VP\left(a\ge0;a\ne1\right)\)
anh Minh đâu r hả cj?