\(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)
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Nàng tiên cá31 tháng 10 2017
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) \(\left(2\frac{1}{2}+3,5\right):\left(-4\frac{1}{6}+3\frac{1}{7}\right)+7,5\)
\(=\left(\frac{5}{2}+\frac{7}{2}\right):\left(\frac{-25}{6}+\frac{22}{7}\right)+\frac{15}{2}\)
\(=6:\frac{-43}{42}+\frac{15}{2}\)
\(=\frac{-72}{43}+\frac{15}{2}\)
\(=\frac{501}{86}\)
14 tháng 5 2022
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: \(=\left(\dfrac{35-32}{60}\right)^2+\dfrac{4}{5}\cdot\dfrac{70-45}{80}\)
\(=\dfrac{1}{400}+\dfrac{4\cdot25}{400}=\dfrac{101}{400}\)
S
25 tháng 3 2023
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right)\cdot\dfrac{1}{3}-\dfrac{2}{7}\cdot\dfrac{49}{4}\)
\(=\dfrac{1}{8}\cdot\dfrac{1}{3}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{7}{2}\)
\(=\dfrac{1}{24}-\dfrac{84}{24}\)
\(=-\dfrac{83}{24}\)
#データネ
NH
25 tháng 3 2023
`(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`
28 tháng 8 2023
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
4 tháng 5 2017
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 \(S=\left\{-23;\dfrac{1}{2}\right\}\)
Giải:
\(\left|-x+\dfrac{2}{5}\right|+\dfrac{1}{2}=3,5\)
\(\Leftrightarrow\left|-x+\dfrac{2}{5}\right|=3,5-\dfrac{1}{2}=3\)
Ta xét các trường hợp:
* Trường hợp 1:
\(-x+\dfrac{2}{5}=3\)
\(\Rightarrow-x=3-\dfrac{2}{5}=2,6\)
\(\Leftrightarrow x=-2,6\)
* Trường hợp 2:
\(-x+\dfrac{2}{5}=-3\)
\(\Rightarrow-x=-3-\dfrac{2}{5}=-3,4\)
\(\Leftrightarrow x=3,4\)
Vậy \(x\in\left\{-2,6;3,4\right\}\)
Chúc bạn học tốt!
$|-x+\dfrac{2}{5}|+\dfrac{1}{2}=3,5$
$=>|-x+0,4|+0,5=3,5$
$=>|-x+0,4|=3,5-0,5=3$
\(=>\left[{}\begin{matrix}-x+0,4=3\\-x+0,4=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}-x=3-0,4=2,6\\-x=-3-0,4=-3,4\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=-2,6\\x=3,4\end{matrix}\right.\)