Giải chi tiết hộ mk:
Rút gọn biểu thức:
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
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24 tháng 7 2019
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
TN
18 tháng 8 2015
C=\(\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)}\right).\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(\sqrt{x}+2\right).\left(x-1\right)-\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x\sqrt{x}-\sqrt{x}+2x-2-\left(x-1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{x-1+x\sqrt{x}-\sqrt{x}}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{\left(x-1\right).\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(x-1\right)}.\frac{\sqrt{x}+1}{\sqrt{x}}\)
C=\(\frac{1}{\sqrt{x}}=\frac{\sqrt{x}}{x}\)
18 tháng 5 2019
ĐKXĐ:\(x\ge0,x\ne1,x\ne\frac{1}{2}\)
a) \(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}=\left[\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+3}{\sqrt{x}-1}\right].\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}+3}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{2\sqrt{x}-1}=\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
b) Ta có \(B< 0\Leftrightarrow\frac{2\sqrt{x}+3}{2\sqrt{x}-1}< 0\)(1)
Vì \(2\sqrt{x}+3>0\)
(1)\(\Leftrightarrow2\sqrt{x}-1< 0\Leftrightarrow\sqrt{x}< \frac{1}{2}\Leftrightarrow x< \frac{1}{4}\)
Kết hợp với ĐK, vậy \(0\le x< \frac{1}{4}\) thì B<0
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}\right).\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(\frac{3x\sqrt{x}+2x}{2x\sqrt{x}+x+\sqrt{x}-1}\)
e moi lop 6