bài 1: tính:
a,1/5-1/2+/-1/4/
b,/-1/4/-3+3/4
c,3/4- 1 2 trên 3 (hỗn số)+/-1 1 trên 6/-2
d,(11/12-15/3+1/12-1/3):2/3
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`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
bn ơi ko thì bn làm từng phép một cũng đc nhé các bn giải hộ mình mai mình phải nộp rồi
`4/7+4`
`=4/7+4/1`
`=4/7+28/7`
`=32/7`
__
`3+6/11`
`=33/11+6/11`
`=39/11`
__
`3-5/7`
`=3/1-5/7`
`=21/7-5/7`
`=16/7`
__
`21/9-2`
`=21/9-18/9`
`=3/9`
`=1/3`
__
`15/24+2`
`=15/24+48/24`
`=63/24`
`=21/16`
__
`63/45-20/25`
`=63/45-4/5`
`=63/45-36/45`
`=27/45`
`=9/15`
__
`3/4-2/8`
`=3/4-1/4`
`=2/4`
__
`6/7-5/8`
`=48/56-35/56`
`=13/56`
__
`37/45-5/9`
`=37/45-25/45`
`=12/45`
`=4/15`
__
`46/39-11/13`
`=46/39-33/39`
`=13/39`
`=1/2`
__
`5/12+3/4+1/3`
`=5/12+9/12+4/12`
`=14/12+4/12`
`=18/12`
`=3/2`
__
`1/2+3/7+11/14`
`=7/14+6/14+11/14`
`=13/14+11/14`
`=24/14`
`=12/7`
__
`7/10-(1/5+1/4)`
`=7/10-(4/20+5/20)`
`=7/10-9/20`
`=14/20-9/20`
`=5/20`
`=1/4`
__
`15/4-2/3-3/4`
`=(15/4-3/4)-2/3`
`=12/4-2/3`
`=3-2/3`
`=9/3-2/3`
`=7/3`
\(a,1\dfrac{4}{7}.3\dfrac{4}{11}.3\dfrac{11}{15}.5\dfrac{5}{8}\)
\(=\dfrac{11}{7}.\dfrac{27}{11}.\dfrac{56}{15}.\dfrac{45}{8}\)
\(=\dfrac{11.27.56.45}{7.11.15.8}\)
\(=\dfrac{1.3.7.3}{1.1.1.1}\)
\(=63\)
\(b,\dfrac{3}{4}.1\dfrac{1}{2}+\dfrac{3}{4}.\dfrac{1}{2}\)
\(=\dfrac{3}{4}.\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(=\dfrac{3}{4}.2\)
\(=\dfrac{3}{2}\)
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)