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\(a,\dfrac{1}{5}-\dfrac{1}{5}:x=\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{1}{5}-\dfrac{3}{5}\\ \Rightarrow\dfrac{1}{5}:x=\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{1}{5}:\dfrac{-2}{5}\\ \Rightarrow x=\dfrac{-1}{2}\)
câu b thiếu đề
a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
\(\frac{1}{2}.x\)+\(\frac{3}{5}\).( x - 2 ) = 3
1/2 . x + 3/5 . x - 3/5 . 2 = 3
x . ( 1/2 + 3/5 ) - 6/5 = 3
x . 11/10 - 6/5 = 3
x . 11/10 = 3 + 6/5
x . 11/10. = 21/5
x. = 21/5 : 11/10
x. = 42/11
Vậy x = 42/11
a, \(\frac{x}{5}-\frac{2}{3}+2x=\frac{1}{2}\)
\(\frac{6x}{30}-\frac{20}{30}+\frac{60x}{30}=\frac{15}{30}\)
\(\frac{66x-20}{30}=\frac{15}{30}\)
\(\Rightarrow\) 66x - 20 = 15
66x = 15 + 20
66x = 35
x = \(\frac{35}{66}\)
Vậy x = \(\frac{35}{66}\)
b, \(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{7}\)
\(\frac{5}{2}-1+3x=\frac{1}{7}\)
\(\frac{3}{2}+3x=\frac{1}{7}\)
3x = \(\frac{1}{7}-\frac{3}{2}\)
3x = \(\frac{-19}{14}\)
x = \(\frac{-19}{42}\)
Vậy x = \(\frac{-19}{42}\)
Phần c lỗi, chắc như thế này
c, \(4\cdot\left(\frac{1}{2}-x\right)+\frac{1}{2}=\frac{-5}{6}+x\)
\(2-4x+\frac{1}{2}=\frac{-5}{6}+x\)
\(\frac{5}{2}-4x=\frac{-5}{6}+x\)
\(-4x-x=\frac{-5}{6}-\frac{5}{2}\)
\(-5x=\frac{-10}{3}\)
x = \(\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
Chúc bn học tốt
3-2(x+2)=5-x
\(\Rightarrow3-2x-4=5-x\)
\(\Rightarrow-2x+x=5+4-3\)
\(\Rightarrow-x=6\)
\(\Rightarrow x=-6\)
3 - 2 . ( x + 2 ) = 5 - x
=> 3 - 2x - 4 = 5 - x
=> -1 - 2x = 5 - x
=> -1 - 5 = - x + 2x
=> x = -6
Vậy x = -6
\(2\left(x-3\right)-5\left(x-4\right)=-7+14.\left(-3\right)\)
\(2x-6-5x+20=-7+\left(-42\right)\)
\(\left(2x-5x\right)+\left(-6+20\right)=-49\)
\(-3x+14=-49\)
\(-3x=-63\)
\(x=21\)
(x - 1)² + 3² = 5².1¹⁰⁰
(x - 1)² + 9 = 25 .1
(x - 1)² + 9 = 25
(x - 1)² = 25 - 9
(x - 1)² = 16
x - 1 = 4 hoặc x - 1 = -4
*) x - 1 = 4
x = 4 + 1
x = 5
*) x - 1 = -4
x = -4 + 1
x = -3
Vậy x = -3; x = 5
\(\left(x+5\right)^3=\left(x+5\right)^2\)
`(x+5)^3 -(x+5)^2 =0`
`(x+5)^2 (x+5-1)=0`
`(x+5)^2 (x+4)=0`
\(=>\left[{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)