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a, Ta có : \(B=\frac{\sqrt{x}-4}{x-2\sqrt{x}}+\frac{3}{\sqrt{x}-2}=\frac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\frac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\frac{B}{A}\Rightarrow P=\frac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{-\sqrt{x}\left(\sqrt{x}-2\right)}{4}=1-\sqrt{x}\)
b, Ta có : \(M=P.\frac{1-\sqrt{x}}{\sqrt{x}-3}\Rightarrow M=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-3}\ge0\)
\(\Rightarrow\sqrt{x}-3\ge0\Leftrightarrow\sqrt{x}\ge3\Leftrightarrow x\ge9\)vì \(\left(\sqrt{x}-1\right)^2\ge0\)
N=\(\left(\frac{x\sqrt{x}+3\sqrt{3}}{x-\sqrt{3x}+3}-2\sqrt{x}\right).\left(\frac{\sqrt{x}+\sqrt{3}}{3-x}\right)\)
ĐKXĐ \(\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\3-x\ne0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\sqrt{3x}+3\ne0\\x\ne3\\x\ge0\end{cases}}\)
\(=\left[\frac{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}{x-\sqrt{3x}+3}-2\sqrt{x}\right].\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\left(\sqrt{x}+\sqrt{3}-2\sqrt{x}\right).\frac{\sqrt{x}+\sqrt{3}}{3-x}\)
\(=\frac{x-2x+3}{3-x}=\frac{3-x}{3-x}=1\)
câu 2 ra |a-b| nha bn mik đăng rồi nhưng bị lỗi nên nó ko hiện lên
B=\(\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{10\sqrt{x}-2x+2\sqrt{x}+3\sqrt{x}-3-x-4\sqrt{x}-\sqrt{x}-4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{-3x+3\sqrt{x}+7\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=\(\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)=\(\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)
Vậy...
\(B=\frac{10\sqrt{x}}{x+3\sqrt{x}-4}-\frac{2\sqrt{x-3}}{\sqrt{x}+4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\)( \(x\ge0;x\ne1\)
=>\(B=\frac{10\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}-\frac{2\sqrt{x}-3}{\sqrt{x}+4}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
=> \(B=\frac{10\sqrt{x}-\left(2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{10\sqrt{x}-\left(2x-5\sqrt{x}+3\right)-\left(x+5\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{-3x+10\sqrt{x}-7}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\)
=> \(B=\frac{\left(\sqrt{x}-1\right)\left(7-3\sqrt{x}\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}=\frac{7-3\sqrt{x}}{\sqrt{x}+4}\)( zì \(x\ge0,x\ne1\)
\(\sqrt{x-1-2\sqrt{x-1}+1}\)+\(\sqrt{x-1+4\sqrt{x-1}+4}\) (\(x\ge1\)
=\(\left|\sqrt{x-1}-1\right|+\left|\sqrt{x-1}-2\right|\)
dat \(\sqrt{x-1}=t\left(t\ge0\right)\)
ta co \(\left|t-1\right|+\left|t-2\right|\)
t |t-1| |t-2| 1 2 0 0 + - - +
nenta co voi0<= t<1 \(1-t+2-t=3-t=3-2\sqrt{x-1}\)
voi 1\(\le t\le2\) \(t-1+2-t=3\)
voi t>2 \(t-1+t-2=2t-3=2\sqrt{x-1}-3\)
b,\(\sqrt{x-4-4\sqrt{x-4}+4}\) =\(\left|\sqrt{x-4}-2\right|\)