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a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}=\dfrac{2}{\sqrt{x}+2}\)
b: A=1/2
=>\(\sqrt{x}+2=4\)
=>\(\sqrt{x}=2\)
=>x=4(loại)
\(A=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)
\(A=\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)\(\div\left(\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(A=\left(\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right):\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{2x+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{4\sqrt{x}}\)
\(A=\frac{2x+1}{4\sqrt{x}}\)
c, \(A=\frac{2x+1}{4\sqrt{x}}=\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\)
ap dụng cô si ta có \(\frac{\sqrt{x}}{2}+\frac{1}{4\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{2}\cdot\frac{1}{4\sqrt{x}}}=\frac{\sqrt{2}}{2}\)
dấu = xảy ra khi \(\frac{\sqrt{x}}{2}=\frac{1}{4\sqrt{x}}\Leftrightarrow x=\frac{1}{2}\) (tm)
1)đặt nhân tử chung quy đồng là xong
2)phân tích x+2cănx-3=(1-cănx)(3+cănx)
3)2a+căn a đặt căn a ra r rút gọn