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\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}-\frac{2\sqrt{x}-1}{\sqrt{x}+2}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}-\frac{\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{2\left(x-2\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-\left(2\sqrt{x}-1\right)\left(x-4\right)}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{x+2\sqrt{x}}{\left(x-4\right)\left(\sqrt{x}+2\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\left[\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(\sqrt{x+2}\right)}\right]:\frac{\sqrt{x}}{\sqrt{x}-2}\)
\(A=\frac{\sqrt{x}}{x-4}\cdot\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)
\(A=\frac{\sqrt{x}-2}{x-4}\)
K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)
Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)
Với x>1 thoả mãn yêu cầu.
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) \(A=\frac{3+\sqrt{9x-3}}{x+\sqrt{x-2}}+\frac{\sqrt{x+1}}{\sqrt{x+2}}+\frac{\sqrt{x-2}}{1-\sqrt{x}}.\)
\(A=\frac{3x+3\sqrt{x-3}}{\left(\sqrt{x-1}\sqrt{x-2}\right)}-\frac{\sqrt{x+1}}{\sqrt{x+2}}-\frac{\sqrt{x-2}}{\sqrt{x-1}}\)
\(A=\frac{3x+3\sqrt{x-3}-\left(\sqrt{x+1}\right)\left(\sqrt{x-1}\right)-\left(\sqrt{x+2}\right)\left(\sqrt{x-2}\right)}{\left(\sqrt{x-1}\right)\left(\sqrt{x+2}\right)}\)
\(A=\frac{3x+3\sqrt{x-3}-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x-1}\right)\left(\sqrt{x+2}\right)}\)
\(A=\frac{3x+3\sqrt{x-3}-x+1-x+4}{\left(\sqrt{x-1}\right)\left(\sqrt{x+2}\right)}=\frac{x+3\sqrt{x+2}}{\left(\sqrt{x-1}\right)\left(\sqrt{x+2}\right)}=\frac{\left(\sqrt{x+1}\right)\left(\sqrt{x+2}\right)}{\left(\sqrt{x+1}\right)\left(\sqrt{x-2}\right)}=\frac{\sqrt{x+1}}{\sqrt{x-1}}\)
Thêm phần
\(ĐK:\hept{\begin{cases}x>0\\\frac{x}{0}\end{cases}}\)tức là x khác 0