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\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)
\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)
\(\Leftrightarrow\frac{38}{x+3}=16\)
\(\Leftrightarrow x+3=2,375\)
\(\Leftrightarrow x=-0,625\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)
\(\Leftrightarrow\frac{38}{x+3}=14\)
\(\Leftrightarrow\left(x+3\right)14=38\)
\(\Leftrightarrow14x+42=38\)
\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)
Vậy \(x=-\frac{2}{7}\)
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
2:(1/2-2/3)^3=2:(-1/6)^3=2.(-6)^3=2.(-216)=-432
\(2:\left(\frac{1}{2}-\frac{3}{4}\right)^3\\ =2:\left(-\frac{1}{4}\right)^3\\ =2\cdot\left(-4\right)^3\\ =-128\)