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\(P=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)
\(=2+\dfrac{2x+2}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
Sửa: \(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}}\)
\(C=\dfrac{\sqrt{x}+3\sqrt{x}+2-x+3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}{\sqrt{x}}\\ C=\dfrac{6\sqrt{x}}{\sqrt{x}}=6\)
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
Ta có : A = \(\left(\frac{x+2}{x.\sqrt{x}-1}+\frac{\sqrt{x}+2}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
= \(\frac{x+2+x+\sqrt{x}-2-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
= \(\frac{x-1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+1}=1\)
Vậy A = 1
ĐKXĐ: \(x>0;x\ne1\)
\(C=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(C=\frac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{x}\)