\(-\frac{3}{5}.x=\frac{21}{10}\)
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a)\(\frac{5}{21}\)+\(\frac{-3}{7}\)<\(\frac{x}{21}\)<\(\frac{-2}{7}\)+\(\frac{8}{21}\)
\(\Rightarrow\)\(\frac{-4}{21}\)<\(\frac{x}{21}\)<\(\frac{2}{21}\)
\(\Rightarrow\)\(\frac{x}{21}\)\(\in\)\(\left\{\frac{-3}{21};\frac{-2}{21};\frac{-1}{21};\frac{0}{21};\frac{1}{21}\right\}\)
vậy x\(\in\)\(\left\{-3;-2;-1;0;1\right\}\)
\(\frac{-3}{5}x=\frac{21}{10}\Rightarrow x=\frac{21}{10}\div\frac{-3}{5}=\frac{21}{10}.\frac{-5}{3}=\frac{7}{2}\)
\(\frac{x}{20}=\frac{4}{5}\Leftrightarrow5x=20.4=80\Leftrightarrow x=80\div5=16\)
P/s : Dễ
a) $\frac{2}{3} - \frac{1}{3} = \frac{{2 - 1}}{3} = \frac{1}{3}$
b) $\frac{7}{{12}} - \frac{5}{{12}} = \frac{{7 - 5}}{{12}} = \frac{2}{{12}} = \frac{1}{6}$
c) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
a,\(\frac{-3}{5}-\left(\frac{-2}{5}+2\right)=\frac{-3}{5}-\frac{8}{5}=1\)
b,\(\frac{-21}{10}+\frac{21}{10}.\frac{5}{7}=-\frac{21}{10}+\frac{105}{70}=-\frac{21}{10}+\frac{15}{10}=-\frac{6}{10}\)
Câu hỏi gì mà ngộ vậy
\(\text{Tìm x , biết :}\)
\(-\frac{3}{5}\cdot x=\frac{21}{10}\)
\(\Rightarrow x=\frac{21}{10}\div-\frac{3}{5}\)
\(\Rightarrow x=\frac{21}{10}\cdot-\frac{5}{3}\)
\(\Rightarrow x=\frac{21\cdot(-5)}{10\cdot3}=-\frac{105}{30}=-\frac{7}{2}\)
\(\text{Vậy x }=-\frac{7}{2}\)
Mình không chắc nữa nhé