\(\frac{\frac{-2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}+2}-\frac{\frac{-2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{2}{4}+2}\)
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Ta có \(\frac{-\frac{2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\frac{2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{-\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{23}{25}.\left(-1\right)\)
\(=\frac{-23}{25}\)
giải:
ta có :
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)
\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
\(=\frac{-1\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}+2}-\frac{-1\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=-1-\left(-1\right)\)
\(=-1+1\)
\(=0\)
\(=\frac{-\left(\frac{2}{3}+\frac{3}{4}-2\right)}{\frac{2}{3}+\frac{3}{4}-2}-\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\left(-1\right)-\left(-1\right)\)
\(=0\)