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\(A=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\left[\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2x+\sqrt{x}-1-x-\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\frac{\sqrt{x}\left(x-2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\frac{\sqrt{x}\left(x-2\right)\left(\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(A=\frac{\sqrt{x}\left(x-2\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{2x\sqrt{x}-\sqrt{x}+x}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
f: Ta có: \(\left(x+1\right)\left(x-2\right)-\left(2-x\right)\left(3-x\right)>0\)
\(\Leftrightarrow x^2-2x+x-2-\left(x-2\right)\left(x-3\right)>0\)
\(\Leftrightarrow x^2-x-2-x^2+5x-6>0\)
\(\Leftrightarrow4x>8\)
hay x>2
g: Ta có: \(\left(2x-1\right)^2\le2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-4x+1-2x^2+4x-2\le0\)
\(\Leftrightarrow2x^2\le1\)
\(\Leftrightarrow x^2\le\dfrac{1}{2}\)
\(\Leftrightarrow-\dfrac{\sqrt{2}}{2}\le x\le\dfrac{\sqrt{2}}{2}\)
a. rút gọn b. Tính giá trị A khi x =\(\sqrt{3+\sqrt{8}}\)
c. Tìm x=\(\sqrt{5}\)
