\(\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
a)Rút gọn
b)Tìm GTNN của M
K
Khách
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24 tháng 7 2019
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
21 tháng 9 2020
a) đk: \(x\ge0;x\ne\left\{\frac{1}{4};1\right\}\)
\(P=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\frac{x+\sqrt{x}}{x-1}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\left[\frac{\left(2x+\sqrt{x}-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{x-1}\right]\cdot\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{\left(x-1\right)\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}}{2\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
21 tháng 9 2020
b) Ta có:
\(P=\frac{x+\sqrt{x}}{x+\sqrt{x}+1}=\frac{\left(x+\sqrt{x}+1\right)-1}{x+\sqrt{x}+1}=1-\frac{1}{x+\sqrt{x}+1}\)
Mà \(x+\sqrt{x}\ge0\left(\forall x\right)\)
\(\Leftrightarrow x+\sqrt{x}+1\ge1\left(\forall x\right)\)
\(\Leftrightarrow\frac{1}{x+\sqrt{x}+1}\le1\left(\forall x\right)\)
\(\Leftrightarrow P=1-\frac{1}{x+\sqrt{x}+1}\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(x+\sqrt{x}=0\Leftrightarrow x=0\)
Vậy Min(P) = 0 khi x = 0
29 tháng 1 2022
\(M=\dfrac{x-3\sqrt{x}+2+x+3\sqrt{x}+2-2x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\sqrt{x}-2}{3\sqrt{x}-6}\)
\(=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2\left(\sqrt{x}-1\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\left(\sqrt{x}-1\right)}{3\left(\sqrt{x}-2\right)^2}\)