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\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)
\(A=\frac{\sqrt{x}-2}{x-4}\)
a: \(P=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b: \(P=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi x=1/4
\(A=\left(\frac{2+\sqrt{x}}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{2-\sqrt{x}}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) \(:\left(2-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\left(\sqrt{x}+1\right)-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}\right]\)
\(:\left[\frac{2\sqrt{x}+2-\sqrt{x}}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+2+x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right]\) \(:\left[\frac{\sqrt{x}+2}{\sqrt{x}+1}\right]\)
\(A=\left[\frac{\sqrt{x}+x-7-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
;))) tớ nhớ dạng RGBT căn bậc 3 lớp 9 nhì :)))????
\(\left(\frac{2x+1}{\sqrt{x^3}-1}-\frac{\sqrt{x}}{x+\sqrt{x+1}}\right).\left(\frac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\frac{2x+1-\sqrt{x}\left(\sqrt{x-1}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right]\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}.\left(1-2\sqrt{x}+x\right)\)
\(=\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)