0,4 nhân x \(-\)\(\frac{1}{5}\)nhân x =\(\frac{3}{4}\)
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\(-\frac{3}{5}xyz^2\cdot\frac{1}{3}xy\cdot\left(-\frac{1}{4}\right)x^5yz\)
\(=\left(-\frac{3}{5}\cdot\frac{1}{3}\cdot\frac{-1}{4}\right)\left(x\cdot x\cdot x^5\right)\left(y\cdot y\cdot y\right)\left(z^2\cdot z\right)\)
\(=\frac{1}{20}x^7y^3z^3\)
a) \(1\frac{2}{7} = 1 + \frac{2}{7} = \frac{9}{2}\)
\(\begin{array}{l}x:1\frac{2}{7} = - 3,5\\x:\frac{9}{7} = - \frac{7}{2}\\x = - \frac{7}{2}.\frac{9}{7}\\x = - \frac{9}{2}\end{array}\)
b) \(0,4.x - \frac{1}{5}.x = \frac{3}{4}\)
\(\begin{array}{l}\frac{2}{5}.x - \frac{1}{5}.x = \frac{3}{4}\\\left( {\frac{2}{5} - \frac{1}{5}} \right).x = \frac{3}{4}\\\frac{1}{5}.x = \frac{3}{4}\\x = \frac{3}{4}:\frac{1}{5}\\x = \frac{3}{4}.5\\x = \frac{{15}}{4}\end{array}\)
Ta có: \(\frac{1}{2}.x+\frac{3}{5}.\left(x-2\right)=3\)
\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)
\(\Rightarrow\frac{11}{10}x-\frac{6}{5}=3\)
\(\Rightarrow\frac{11}{10}x=3+\frac{6}{5}=\frac{21}{5}\)
\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{42}{11}\)
Vậy \(x=\frac{42}{11}\)
Ủng hộ tớ nha?
\(0,4\cdot x-\frac{1}{5}\cdot x=\frac{3}{4}\)
\(0,4\cdot x-0,2\cdot x=0,75\)
\(\left(0,4-0,2\right)\cdot x=0,75\)
\(0,2\cdot x=0,75\)
\(x=0,75:0,2\)
\(x=3,75\)
x.(0,4-0,2)=0,75
0,2x=0,75
x=3,75