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Bài 5
a, đk x >= 0 ; x khác 1
\(A=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\dfrac{x-2\sqrt{x}+1}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b, đk x >= 0 ; x khác 1
\(B=\left(\dfrac{\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)-\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)
\(=\left(\dfrac{x\sqrt{x}+x+2\sqrt{x}+2+x\sqrt{x}-\sqrt{x}+x-1-x\sqrt{x}-2x-2\sqrt{x}-1}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}-1}{x+\sqrt{1}+x}\)
\(=\dfrac{x\sqrt{x}-\sqrt{x}}{\left(x-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x-1\right)}{\left(x-1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)