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1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
\(=10.\dfrac{1}{10}.\dfrac{4}{3}+3.7-\dfrac{1}{6}.2\)
\(=1.\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{28}{21}+\dfrac{441}{21}+\dfrac{7}{21}\)
\(=\dfrac{469}{21}+\dfrac{7}{21}\)
\(=\dfrac{68}{3}\)
\(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+3.\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)
= 10 . 0,1 . \(\dfrac{4}{3}\) + 3. 7 - \(\dfrac{1}{6}.2\)
= 1 . \(\dfrac{4}{3}\) + 21 - \(\dfrac{1}{3}\)
= \(\dfrac{4}{3}+21-\dfrac{1}{3}\)
= 22
a) \(10\sqrt{0,01}.\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\sqrt{\frac{10}{100}}.\sqrt{\frac{4^2}{3^2}}+3.\sqrt{7^2}-\frac{1}{6}\sqrt{2^2}\)
\(=10.\frac{\sqrt{10}}{10}.\frac{4}{3}+3.7-\frac{1}{6}.2\)
\(=\frac{4\sqrt{10}}{3}+27-\frac{1}{3}\)
\(=\frac{4}{3}\sqrt{10}+\frac{80}{3}\)
b) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(0,8-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\left(\frac{4}{5}-\frac{3}{4}\right)^2\)
\(=\frac{17}{12}.\left(\frac{1}{20}\right)^2\)
\(=\frac{17}{12}.\frac{1}{400}\)
\(=\frac{17}{4800}\)
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) \(\left(x-\frac{2}{5}\right)\left(x+\frac{3}{7}\right)=0\)
=> \(\orbr{\begin{cases}x-\frac{2}{5}=0\\x+\frac{3}{7}=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{3}{7}\end{cases}}\)
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
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
\(10.\sqrt{0,01}.\sqrt{\frac{16}{9}}+3\sqrt{19}-\frac{1}{6}\sqrt{4}\)
\(=10.\sqrt{\left(\frac{1}{10}\right)^2}.\sqrt{\left(\frac{4}{3}\right)^2}+3\sqrt{19}-\frac{1}{6}\sqrt{2^2}\)
\(=10.\frac{1}{10}.\frac{4}{3}+3\sqrt{19}-\frac{1}{6}.2\)
\(=\frac{4}{3}+3\sqrt{19}-\frac{1}{3}\)
\(=\left(\frac{4}{3}-\frac{1}{3}\right)+3\sqrt{19}\)
\(=1+3\sqrt{19}\)