ĐKXĐ : \(x\ge0\)
\(\frac{2\sqrt{x}-4}{x-4}=\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{2}{\sqrt{x}+2}\le\frac{2}{2}=1\)
Dấu = xảy ra \(\Leftrightarrow x=0\)
Cho x\(\in [0;1]\)
Tìm Max của P=\(13\sqrt{x^2-x^4}\ + 9\sqrt{x^2+x^4}\)
tìm Max: 1) \(\dfrac{2\sqrt{x}}{x+1}\) ; 2)\(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)
\(\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{4x}{x-4}-\frac{2-\sqrt{x}}{2+\sqrt{x}}\right):\frac{4\sqrt{x^3-12\sqrt{x}}}{2x-\sqrt{x^3}}\)
\((\frac{\sqrt{x}}{\sqrt{x}-4}+\frac{4}{\sqrt{x}+4}):\frac{x+16}{\sqrt{x}+2}\)
\(\frac{\sqrt{x}}{\sqrt{x}-2}-\left(\frac{2}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}\right)\)
chứng minh rằng
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}=\frac{2}{\sqrt[]{x}}\)
\(\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}+2}+\frac{x-4}{2-\sqrt{x}}\right)\)+\(\sqrt{x}\)
\(1>A=\frac{a\sqrt{a}+1}{a-\sqrt{a}+1}-\frac{2a-2\sqrt{a}}{\sqrt{a}-1}\)
\(2>C=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{2+2}}\right)\cdot\frac{4-x}{2\sqrt{x}}\)
3> \(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
4> \(E=\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}\)
tìm đk để các bt trên có nghĩa và rút gọn chúng.
mình đg cần trg ngày, thx nhìu
cho A = \(\left(\frac{4\sqrt{x}}{\sqrt{x}+2}-\frac{8x}{x-4}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\) đk x > 0, x khác 4
a, rút gọn A
b, tìm x đề A = -1