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\(a,3-\left(17-x\right)=-12\\ \Rightarrow17-x=15\\ \Rightarrow x=2\\ b,-26-\left(x-7\right)=0\\ \Rightarrow x-7=-26\\ \Rightarrow x=-19\\ c,25+\left(-2+x\right)=5\\ \Rightarrow-2+x=20\\ \Rightarrow x=18\\ d,30+\left(32-x\right)=10\\ \Rightarrow32-x=-20\\ \Rightarrow x=52\)
a) `3-(17-x)=-12`
`3-17+x=-12`
`x=-12-3+17`
`x=2`
b) `-26-(x-7)=0`
`-26-x+7=0`
`-19-x=0`
`x=-19`
c) `25+(-2+x)=5`
`25-2+x=5`
`x=5-25+2`
`x=-18`
d) `30+(32-x)=10`
`30+32-x=10`
`62-x=10`
`x=52`
3 - (17 - x) = -12
(17 - x) = -12 - 3
(17 - x) = -15
x = 17 + -15
x = 2
25 + (-2 + x) = 5
25 - 2 + x = 5
23 + x = 5
x = 5 - 23
x = -18
-25 - (x - 7) = 0
- (x - 7) = 0 + -25
- (x - 7) = -25
x = -25 + 7
x = -18
30 + (32 + x) = 10
62 + x = 10
x = 62 - 10
x = 52
a) 3 - ( 17 - x ) = -12
3 - 17 - x = -12
- x = 2
=> x = -2
b) 25 + ( -2 +x ) = 5
25 - 2 + x = 5
=> x = -18
c) -25 - ( x - 7 ) = 0
-25 - x + 7 = 0
- x = 18
=> x = -18
30 + ( 32 - x ) = 10
30 + 32 - x = 10
-x = -52
=> x = 52
Chúc em học tốt!!!
a,(2x+7)+135=0 b, 1/2x-2/5=1/5
2x+7=0-135 1/2x=1/5+2/5
2x+7=-135 1/2x=3/5
2x=-135-7 x=3/5:1/2
2x=-142 x=6/5
x=-142:2 Vậy x=6/5
x=-71
Vậy x=-71
c, 10-|x+1|=5 d, 1/2x+150%x=2014
|x+1|=10-5 2x=2014
|x+1|=5 x=2014:2
*TH1:x+1=5 *TH2:x+1=-5 x=1007
x=5-1 x=-5-1 Vậy x=1007
x=4 x=-6
Vậy x=4 hoặc x=-6
Bài 2:
a: Ta có: \(2x+79=x+45\)
nên 2x-x=45-79
hay x=-24
b: Ta có: \(15-\left(x+7\right)=\left(x-8\right)+22\)
\(\Leftrightarrow8-x-x-14=0\)
\(\Leftrightarrow2x=22\)
hay x=11
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)
\(\Leftrightarrow x-\dfrac{1}{3}=3\)
\(\Leftrightarrow x=3+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)
\(\Leftrightarrow x=\dfrac{10}{3}\)
\(b,x+30\%x=-1,31\)
\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)
\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)
\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)
\(\Leftrightarrow x=-\dfrac{131}{130}\)
\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)
\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)
\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)
\(\Leftrightarrow x=\dfrac{3}{20}\)
a, 3 - (17 -x ) = -12
3- 17+ x = 12
-14 = 12 - x
12 - (-14) = x
2 = x
x = 2
a;x=2 b;x=35 c;x=-18 d;x=52