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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`
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, 3 - (17 -x ) = -12
3- 17+ x = 12
-14 = 12 - x
12 - (-14) = x
2 = x
x = 2
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) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
a) x+7=-12
x=(-12)-7
x=-19
b)x-15=-21
x=(-21)+15
x=-6
c)13-x=20
x=13-20
x=-7
d)17-(2+x)=3
x=17-3
x=14
x=14-2
x=12
a)2
b)26
c) -243
d)52
a, => 3-17+x=-12
=> x-14=-12
=> x=-12+14 = 2
b, => 26-x+7=0
=> 33-x=0
=> x=33-0=33
c, => 250-2+x=5
=> x+248=5
=> x=5-248 = -243
d, => 30+32-x=10
=> 62-x=10
=> x=62-10=52
Tk mk nha