Tìm x :
\(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
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VH
3
CW
9 tháng 7 2016
\(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{3}+\frac{1}{6}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{2}\)
\(\sqrt{x}=\frac{1}{2}:\frac{5}{12}\)
\(\sqrt{x}=\frac{6}{5}\)
\(\sqrt{x}=\sqrt{\frac{36}{25}}\)
\(x=\frac{36}{25}\)
PT
1
21 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{243}-\frac{1}{2}\sqrt{12}-2\sqrt{75}+\sqrt{27}\)
\(=\sqrt{3}\cdot9-\frac{1}{2}\cdot\sqrt{3}\cdot2-2\cdot\sqrt{3}\cdot5+\sqrt{3}\cdot3\)
\(=\sqrt{3}\left(9-1-10+3\right)\)
\(=\sqrt{3}\cdot1=\sqrt{3}\)
b) Ta có: \(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{5}{1+\sqrt{6}}-6\sqrt{\frac{1}{6}}\)
\(=\frac{\left(2\sqrt{3}-3\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\cdot\left(\sqrt{3}+\sqrt{2}\right)}+\frac{5\cdot\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\sqrt{36\cdot\frac{1}{6}}\)
\(=-\sqrt{6}+\frac{5\left(\sqrt{6}-1\right)}{5}-\sqrt{6}\)
\(=-2\sqrt{6}+\sqrt{6}-1\)
\(=-\sqrt{6}-1\)
Bài 2: Rút gọn
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
\(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
\(\Leftrightarrow\frac{5}{12}\sqrt{x}=\frac{1}{3}+\frac{1}{6}\)
\(\Leftrightarrow\frac{5}{12}\sqrt{x}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{2}:\frac{5}{12}\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{5}\)
\(\Leftrightarrow x=\left(\frac{6}{5}\right)^2\)
\(\Leftrightarrow x=\frac{36}{25}\)