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nhân chéo là đc:
3(x+2)=-4(x-5)
3x+6=-4x+20
3x+4x=20-6
7x =14
x =2
Vậy x=2
\(\frac{1}{2}x-\frac{1}{4}=\frac{-1}{2}\)
\(\frac{1}{2}x=\frac{-1}{2}+\frac{1}{4}\)
\(\frac{1}{2}x=\frac{-2+1}{4}\)
\(\frac{1}{2}x=\frac{-1}{4}\)
\(x=\frac{-1}{4}:\frac{1}{2}\)
\(x=\frac{-1}{4}.2\)
\(x=\frac{-1}{2}\)
Vậy .....................
\(\frac{1}{2}x-\frac{1}{4}=-\frac{1}{2}\)
\(\frac{1}{2}x=-\frac{1}{2}+\frac{1}{4}\)
\(\frac{1}{2}x=-\frac{2}{4}+\frac{1}{4}\)
\(\frac{1}{2}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{1}{2}\)
\(x=-\frac{1}{4}\cdot2\)
\(x=-\frac{1}{2}\)
Vậy .....
\(\frac{x+3}{-4}=-\frac{9}{x+3}\)
\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x+3\right)^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)
Vậy ....
quy đồng
\(\left(x+3\right)^2=36\)
\(\left(x+3\right)^2-6^2=0\)
áp dụng định lí " \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được
\(\left(x+3-6\right)\left(x+3+6\right)=0\)
\(x=3,x=-9\)
b) \(\left(2,4.x-36\right)\div1\frac{5}{7}=-1\)
\(\left(2,4.x-36\right)=-1.\frac{12}{7}\)
\(2,4.x-36=-\frac{12}{7}\)
\(2,4.x=-\frac{12}{7}+36\)
\(2,4.x=\frac{240}{7}\)
\(x=\frac{240}{7}\div2,4\)
\(x=\frac{100}{7}\)
a) \(\frac{3}{x-5}=\frac{-4}{x+2}\)(ĐKXĐ: \(x\ne5;x\ne-2\))
\(\Rightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)(thỏa mãn ĐKXĐ)
b) \(2,4x-36=-\frac{7}{12}\)
\(\Leftrightarrow2,4x=\frac{425}{12}\)
\(\Leftrightarrow x=\frac{2125}{144}\)
c) \(\left(\frac{19}{5}-2x\right).\frac{4}{3}=\frac{40}{7}\)
\(\Leftrightarrow\frac{19}{5}-2x=\frac{30}{7}\)
\(\Leftrightarrow2x=-\frac{17}{35}\)
\(\Leftrightarrow x=-\frac{17}{70}\)
= 10/3.x+ 67/4=-53/4
10/3.x=-53/4-67/4
10/3.x=-30
x=-30:10/3
x=-9
\(\frac{3}{5}+\frac{-1}{25}-\frac{7}{20}\)
\(=\frac{3}{5}-\frac{1}{25}-\frac{7}{20}\)
\(=\frac{60}{100}-\frac{4}{100}-\frac{35}{100}\)
\(=\frac{21}{100}\)
\(-52+\frac{2}{3}x=-46\)
\(\frac{2}{3}x=-46+52\)
\(\frac{2}{3}x=6\)
\(x=6:\frac{2}{3}\)
\(x=9\)
\(\frac{3}{4}+\frac{1}{5}:\frac{7}{10}\)
\(=\frac{3}{4}+\frac{2}{7}\)
\(=\frac{29}{28}\)
\(\frac{3}{4}\)+ \(\frac{1}{5}\): \(\frac{7}{10}\)= \(\frac{3}{4}\)+ \(\frac{2}{7}\)
= \(\frac{28}{28}\)= 1