\(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)
1) (-6,5).5,7+5,7.(-3,5)
2) \(\left(1-\dfrac{2}{5}\right)^2\) + \(\dfrac{3}{5}\) + \(\dfrac{-7}{10}\)
chỗ \(\dfrac{3}{5}\) là \(\left|\dfrac{-3}{5}\right|\)
1: \(=5.7\left(-6.5-3.5\right)=-10\cdot5.7=-57\)
Tính :
a)\(\left(2+\dfrac{4}{5}-\dfrac{7}{12}\right).\left(\dfrac{6}{7}-\dfrac{2}{5}\right)^2\)
b) \(\left(2\dfrac{1}{2}+3,5\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7,5\)
a) (\(2+\dfrac{4}{5}-\dfrac{7}{12}\)
Tính
a, \(\left(-3\right)^2+\left(\dfrac{-1}{2}\right)^3-\left(3,5\right)^0\)
b, \(2\dfrac{1}{4}+\dfrac{2}{5}.0,3-\left(\dfrac{12}{5}-3\right)^2\)
a: \(=9-\dfrac{1}{8}-1=8-\dfrac{1}{8}=\dfrac{63}{8}\)
b: \(=2.25+0.4\cdot0.3-\left(-\dfrac{3}{5}\right)^2\)
\(=3.45-0.36=3.09\)
Tính :
a)\(\left(2+\dfrac{4}{5}-\dfrac{7}{12}\right).\left(\dfrac{6}{7}-\dfrac{2}{5}\right)^2\)
b) \(\left(2\dfrac{1}{2}+3,5\right):\left(-4\dfrac{1}{6}+3\dfrac{1}{7}\right)+7,5\)
a) \(\left(2+\frac{4}{5}-\frac{7}{12}\right).\left(\frac{6}{7}-\frac{2}{5}\right)^2\)
\(=\left(\frac{120}{60}+\frac{48}{60}-\frac{70}{60}\right).\left(\frac{36}{49}-2.\frac{6}{7}.\frac{2}{5}+\frac{4}{25}\right)\)
\(=\frac{98}{60}.\left(\frac{36}{49}-\frac{24}{35}+\frac{4}{25}\right)\)
\(=\frac{49}{30}.\left(\frac{900}{1225}-\frac{840}{1225}+\frac{196}{1225}\right)\)
\(=\frac{49}{30}.\frac{256}{1225}\)
\(=\frac{128}{375}\)
b) \(\left(2\frac{1}{2}+3,5\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7,5\)
\(=\left(\frac{5}{2}+\frac{7}{2}\right):\left(\frac{-25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)
\(=6:\frac{-43}{42}+\frac{15}{2}\)
\(=\frac{-72}{43}+\frac{15}{2}\)
\(=\frac{501}{86}\)
1:
\(\dfrac{\dfrac{1}{2}-3,5}{\dfrac{2}{3}-\dfrac{1}{3}.\left(1+\dfrac{1}{4}\right)}\)
2:
a)\(\dfrac{7}{9}.\left(1\dfrac{1}{6}-1:1\dfrac{1}{19}-\dfrac{1}{15}\right)+4\dfrac{2}{5}:24\)
b)\(\left(\dfrac{7}{12}-\dfrac{8}{15}\right)^2+\dfrac{4}{5}.\left(\dfrac{7}{10}-\dfrac{9}{16}\right)\)
Bài 2:
a: \(=\dfrac{7}{9}\left(\dfrac{7}{6}-\dfrac{19}{20}-\dfrac{1}{15}\right)+\dfrac{22}{5}\cdot\dfrac{1}{24}\)
\(=\dfrac{7}{9}\cdot\dfrac{3}{20}+\dfrac{22}{120}=\dfrac{7}{60}+\dfrac{11}{60}=\dfrac{18}{60}=\dfrac{3}{10}\)
b: \(=\left(\dfrac{35-32}{60}\right)^2+\dfrac{4}{5}\cdot\dfrac{70-45}{80}\)
\(=\dfrac{1}{400}+\dfrac{4\cdot25}{400}=\dfrac{101}{400}\)
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).\)\(\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\dfrac{49}{4}\)
\(=\dfrac{1}{8}\cdot\dfrac{1}{3}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{84}{24}\)
\(=-\dfrac{83}{24}\)
#データネ
`(7/8-3/4)xx1/3-2/7xx(3,5)^2`
`=(7/8-6/8)xx1/3-2/7xx12,25`
`=1/8xx1/3-2/7xx12,25`
`=1/24-7/2`
`=1/24-84/24`
`=-83/24`
Tìm x:
a) \(\dfrac{1}{3}.x+\dfrac{2}{5}\left(x-1\right)=0\)
b)\(-5.\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}.\left(x-\dfrac{2}{3}\right)=x\)
c)\(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\)
d)\(9.\left(3x+1\right)^2=16\)
a: =>1/3x+2/5x-2/5=0
=>11/15x-2/5=0
=>11/15x=2/5
=>x=2/5:11/15=2/5*15/11=30/55=6/11
b: =>-5x-1-1/2x+1/3=x
=>-11/2x-2/3-x=0
=>-13/2x=2/3
=>x=-2/3:13/2=-2/3*2/13=-4/39
c: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=1/3 hoặc x=-1/2
d: 9(3x+1)^2=16
=>(3x+1)^2=16/9
=>3x+1=4/3 hoặc 3x+1=-4/3
=>3x=1/3 hoặc 3x=-7/3
=>x=1/9 hoặc x=-7/9
a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)
Giải các phương trình sau :
a) \(\left(4x-10\right)\left(24+5x\right)=0\)
b) \(\left(3,5-7x\right)\left(0,1x+2,3\right)=0\)
c) \(\left(3x-2\right)\left(\dfrac{2\left(x+3\right)}{7}-\dfrac{4x-3}{5}\right)=0\)
d) \(\left(3,3-11x\right)\left(\dfrac{7x+2}{5}+\dfrac{2\left(1-3x\right)}{3}\right)=0\)
a) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}=\dfrac{5}{2}\\x=-\dfrac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{24}{5};\dfrac{5}{2}\right\}\)
b) \(\left(3.5-7x\right)\left(0.1x+2.3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3.5-7x=0\\0.1x+2.3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3.5}{7}=\dfrac{1}{2}\\x=-\dfrac{2.3}{0.1}=-23\end{matrix}\right.\)
Vậy \(S=\left\{-23;\dfrac{1}{2}\right\}\)