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b) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
=> 3.(x + 2) = (x - 5). (-4)
= 3x + 6 = -4x + 20
= 3x - 4x = 26
=> -x = 26
=> x = -26
c) \(\frac{x}{-2}=\frac{-8}{x}\)
=> x.x = -2 . (-8)
=> x2 = 16
=> x = 4
t i c k nha!!! 645646677778879078452352543546456457567564546567
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
3/2-(43/8+X-75/24):50/3=1/2
3/2-(43/8+X-75/24)=25/3
43/8+X-75/24=-41/6
43/8+X=(-41/6)+75/24
43/8+X=(-85/24)
X=85/24-43/8
X=-109/12
Vậy X=-109/12
\(\frac{3}{2}-\left(5\frac{3}{8}+x-\frac{75}{24}\right):16\frac{2}{3}\)\(=\frac{1}{2}\)
\(\left(5\frac{3}{8}+x-\frac{75}{24}\right):16\frac{2}{3}=\frac{3}{2}-\frac{1}{2}\)
\(\left(5\frac{3}{8}+x-\frac{75}{24}\right):16\frac{2}{3}=1\)
\(5\frac{3}{8}+x-\frac{75}{24}\)\(=1.16\frac{2}{3}\)
\(5\frac{3}{8}+x-\frac{75}{24}=\frac{50}{3}\)
\(5\frac{3}{8}+x=\frac{50}{3}+\frac{75}{24}\)
\(5\frac{3}{8}+x=\frac{475}{24}\)
\(x=\frac{475}{24}-5\frac{3}{8}\)
\(x=\frac{173}{12}\)