1b, Cho biểu thức F = \(\left(\frac{1}{x+3\sqrt{x}}-\frac{1}{\sqrt{x}+3}\right):\frac{1-\sqrt{x}}{x+6\sqrt{x}+9}\)
Với x > 0; x # 1
a, Rút gọn F
b, Tìm x để F = \(\frac{5}{2}\)
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23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
NC
29 tháng 8 2020
Bài làm:
Ta có:
\(P=\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\left(\frac{\sqrt{x}-9}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(P=\frac{x-9-x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\left[\frac{\left(9-\sqrt{x}\right)\left(3+\sqrt{x}\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right]\)
\(P=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{-x+6\sqrt{x}+27+x-4\sqrt{x}+2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\div\frac{x+2\sqrt{x}+20}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(P=\frac{3}{\sqrt{x}+3}\cdot\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{x+2\sqrt{x}+20}\)
\(P=\frac{3\left(\sqrt{x}-2\right)}{x+2\sqrt{x}+20}=\frac{3\sqrt{x}-6}{x+2\sqrt{x}+20}\)
NH
2 tháng 10 2019
a, C = \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left[\frac{\sqrt{x}}{3+\sqrt{x}}-\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]:\left[\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right]\)
\(=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(3+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
28 tháng 6 2019
a) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(C=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(C=\frac{3\sqrt{x}-x+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(C=\frac{3\left(\sqrt{x}+3\right)\cdot\sqrt{x}\left(\sqrt{x}-3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)\cdot2\left(\sqrt{x}+2\right)}\)
\(C=\frac{3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
b) Dễ thấy \(C=\frac{3\sqrt{x}}{2\left(\sqrt{x}+2\right)}\ge0\forall x\)do đó không có giá trị của x thỏa mãn \(C< -1\)
AN
12 tháng 12 2019
a)\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\frac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}.\left(\sqrt{x}+1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-2}\)
b)\(\frac{1}{M}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=\frac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0,\forall x\ge0\)
\(\Leftrightarrow\sqrt{x}+1\ge1\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}\le1\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+1}\le3\)
\(\Leftrightarrow-\frac{3}{\sqrt{x}+1}\ge-3\)
\(\Leftrightarrow1-\frac{3}{\sqrt{x}+1}\ge-2\)
Dấu "=" xảy ra khi x=0
Vậy \(Min_{\frac{1}{M}}=-2\) khi x=0
