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Nếu \(3\frac{2}{\sqrt{10^2+2^2+40}}\)là hỗn số
=> B = \(2\sqrt{\frac{0,01}{1,21}}+3\frac{2}{\sqrt{10^2+2^2+40}}-\frac{3}{4}\)
B = \(2\sqrt{\frac{1}{121}}+3\frac{2}{144}-\frac{3}{4}\)
B = \(\frac{2}{11}+3\frac{1}{6}-\frac{3}{4}\)
B = \(\frac{2}{11}+\frac{19}{6}-\frac{3}{4}\)
B = \(\frac{343}{132}\)
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a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)
\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)
\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)
Vậy \(A=\dfrac{7}{6}.\)
b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)
\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)
\(=-\dfrac{217}{396}.\)
Vậy \(B=-\dfrac{217}{396}.\)
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a: \(\Leftrightarrow4x+\dfrac{3}{4}=2\cdot\dfrac{2}{5}+0.01\cdot10=\dfrac{9}{10}\)
=>4x=3/20
hay x=3/80
b: \(\Leftrightarrow\left|x\right|=4+\dfrac{1}{8}-9=-\dfrac{39}{8}\)(vô lý)
c: 2x(x-2/3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
d: \(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
=>259-7x=3x+39
=>-10x=-220
hay x=22
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{16}x+\frac{3}{4}=2\sqrt{\frac{4}{25}}+0,01.\sqrt{100}\)
=> \(4x+\frac{3}{4}=2\cdot\frac{2}{5}+0,01\cdot10\)
=> \(4x+\frac{3}{4}=\frac{4}{5}+0,1\)
=> \(4x+\frac{3}{4}=0,9\)
=> \(4x=0,9-\frac{3}{4}\)
=> \(4x=0,15\)
=> \(x=0,15:4=0,0375\)
b) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{16x}+\frac{3}{4}=2\sqrt{\frac{4}{25}}+0,01\cdot\sqrt{100}\)
=> \(\sqrt{16}\cdot\sqrt{x}+\frac{3}{4}=2\cdot\frac{2}{5}+\frac{1}{100}\cdot10\)
=> \(4\cdot\sqrt{x}+\frac{3}{4}=\frac{4}{5}+\frac{1}{10}\cdot1\)
=> \(4\cdot\sqrt{x}+\frac{3}{4}=\frac{4}{5}+\frac{1}{10}\)
=> \(4\cdot\sqrt{x}+\frac{3}{4}=\frac{8}{10}+\frac{1}{10}=\frac{9}{10}\)
=> \(4\cdot\sqrt{x}=\frac{9}{10}-\frac{3}{4}=\frac{3}{20}\)
=> \(\sqrt{x}=\frac{3}{20}:4\)
=> \(\sqrt{x}=\frac{3}{80}\)
=> \(x=\frac{9}{6400}\)
Vậy x = 9/6400
b) \(2\frac{3}{4}x=3\frac{1}{7}:0,01\)
=> \(\frac{11}{4}x=\frac{22}{7}:\frac{1}{100}\)
=> \(\frac{11}{4}x=\frac{22}{7}\cdot100\)
=> \(\frac{11}{4}x=\frac{2200}{7}\)
=> \(x=\frac{2200}{7}:\frac{11}{4}=\frac{2200}{7}\cdot\frac{4}{11}=\frac{800}{7}\)
Vậy x = 800/7
c) \(\left|x\right|+3^2=2^2+\left(\frac{1}{2}\right)^3\)
=> \(\left|x\right|+9=4+\frac{1}{8}\)
=> \(\left|x\right|+9=\frac{33}{8}\)
=> \(\left|x\right|=\frac{33}{8}-9=-\frac{39}{8}\)
Vì \(\left|x\right|\ge0\)mà \(-\frac{39}{8}< 0\)
=> x không thỏa mãn
![](https://rs.olm.vn/images/avt/0.png?1311)