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Bài3: a)(1989x1990+3978):(1992x1991-3984)=. (1989x1990+1989x2):(1992x1991-1992x2)=[1989x(1990+2)]:[1992x(1991-2)]=[1989x1992]:[1992x1989]=38620888:38620888=1
`2/(3.5) + 2/(5.7) + 2/(7.9) + .... + 2/(37.39)`
`=2 . (1/(3.5) + 1/(5.7) + .... + 1/(37.39)`
`=2 .[ 1/2 .(1/3 -1/5 + 1/5 -1/7 +....+1/37 -1/39)]`
`= 1/3 - 1/39`
`=4/13`
Câu 1: \(-7/6+15/6=8/6=4/3\)
Câu 2: \(x=7/12-1/6=>x=7/12-2/12=>x=5/12\) nhé =)
câu 1
\(\dfrac{-7}{6}+\dfrac{15}{6}=\dfrac{-7+15}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)
câu 2
\(x+\dfrac{1}{6}=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}-\dfrac{1}{6}\)
\(x=\dfrac{7}{12}-\dfrac{2}{12}=\dfrac{7-2}{12}\)
\(x=\dfrac{5}{12}\)
\(-\dfrac{3}{5}\times\dfrac{5}{7}+\dfrac{-3}{5}\times\dfrac{4}{7}+\dfrac{-3}{5}\times\dfrac{6}{7}\)
\(=-\dfrac{3}{5}\times\left(\dfrac{5}{7}+\dfrac{4}{7}+\dfrac{6}{7}\right)\)
\(=-\dfrac{3}{5}\times\dfrac{15}{7}=-\dfrac{9}{7}\)
Bài 4:
a: \(\dfrac{x}{15}=\dfrac{2}{3}\)
=>\(x=15\cdot\dfrac{2}{3}=10\)
b: \(x+\dfrac{-8}{13}=\dfrac{-21}{13}\)
=>\(x-\dfrac{8}{13}=-\dfrac{21}{13}\)
=>\(x=-\dfrac{21}{13}+\dfrac{8}{13}=-\dfrac{13}{13}=-1\)
c: \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
=>\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{1}{10}+\dfrac{5}{10}=\dfrac{6}{10}=\dfrac{3}{5}\)
=>\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
d: \(\dfrac{3}{4}+\dfrac{1}{4}:x=-3\)
=>\(\dfrac{1}{4}:x=-3-\dfrac{3}{4}=-\dfrac{15}{4}\)
=>\(x=\dfrac{1}{4}:\dfrac{-15}{4}=\dfrac{-1}{15}\)
e: \(\left(\dfrac{2}{3}-x\right)\left(x-\dfrac{1}{4}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}-x=0\\x-\dfrac{1}{4}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{4}\end{matrix}\right.\)
f: 0,51-x=-0,35
=>x=0,51-(-0,35)
=>x=0,51+0,35=0,86
Bài 3:
a: \(-\dfrac{1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)
\(=\dfrac{-3+5+2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)
\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}=\dfrac{31}{24}\)
c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)\)
\(=\dfrac{3}{5}:\dfrac{7}{20}\)
\(=\dfrac{3}{5}\cdot\dfrac{20}{7}=\dfrac{3\cdot4}{7}=\dfrac{12}{7}\)
d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)
\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}\)
\(=1-\dfrac{1}{8}=\dfrac{7}{8}\)
e: \(\dfrac{-5}{7}+\dfrac{8}{11}+\left(-\dfrac{2}{7}\right)+\dfrac{1}{4}+\dfrac{3}{11}\)
\(=\left(-\dfrac{5}{7}-\dfrac{2}{7}\right)+\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{1}{4}\)
\(=-1+1+\dfrac{1}{4}=\dfrac{1}{4}\)
f: \(\dfrac{2}{7}\cdot\dfrac{5}{13}+\dfrac{2}{7}\cdot\dfrac{6}{13}-\dfrac{2}{7}\cdot1\dfrac{11}{13}\)
\(=\dfrac{2}{7}\cdot\left(\dfrac{5}{13}+\dfrac{6}{13}-1-\dfrac{11}{13}\right)\)
\(=\dfrac{2}{7}\cdot\left(-1\right)=-\dfrac{2}{7}\)
g: \(15,3-21,5-3\cdot1,5\)
=15,3-21,5-4,5
=15,3-26
=-10,7
h: \(2\left(4^2-2\cdot4,1\right)+1.25:5\)
=2(16-8,2)+0,25
=15,6+0,25
=15,85
c) Ta có: \(\dfrac{5}{9}\cdot\dfrac{7}{13}-\dfrac{30}{23}+\dfrac{6}{9}\cdot\dfrac{5}{13}+\dfrac{7}{23}\)
\(=\dfrac{5}{13}\cdot\dfrac{7}{9}+\dfrac{6}{9}\cdot\dfrac{5}{13}-\dfrac{30}{23}+\dfrac{7}{23}\)
\(=\dfrac{5}{13}\left(\dfrac{7}{9}+\dfrac{6}{9}\right)-1\)
\(=\dfrac{5}{13}\cdot\dfrac{13}{9}-1\)
\(=\dfrac{-4}{9}\)
d) Ta có: \(\left(-3.2\right)\cdot\dfrac{-15}{64}+\left(0.8-2\dfrac{4}{5}\right):3\dfrac{2}{3}\)
\(=\dfrac{16}{5}\cdot\dfrac{15}{64}+\left(\dfrac{4}{5}-\dfrac{14}{5}\right):\dfrac{11}{3}\)
\(=\dfrac{3}{4}-2:\dfrac{11}{3}\)
\(=\dfrac{3}{4}-2\cdot\dfrac{3}{11}\)
\(=\dfrac{3}{4}-\dfrac{6}{11}=\dfrac{33}{44}-\dfrac{24}{44}=\dfrac{9}{44}\)