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\(2x+12=3\left(x-7\right)\)
\(\Leftrightarrow2x+12=3x-21\)
\(\Leftrightarrow2x-3x=-12-21\)
\(\Leftrightarrow-x=-33\)
\(\Leftrightarrow x=33\)
bỏ dấu\(\Leftrightarrow\)cx đc nhé,lp 6 chx hok đến phương trình
#hoktot<3#
\(2x+12=3\left(x-7\right)\)
\(\Leftrightarrow2x+12=3x-21\)
\(\Leftrightarrow2x+12-3x+21=0\)
\(\Leftrightarrow-x+33=0\Leftrightarrow33-x=0\Leftrightarrow x=33\)
3 . (-x + 1) - 2.(x + 7) - 32 = -(-2x - 9) - 158
=> -3x + 3 - 2x - 14 - 9 = 2x + 9 - 1
=> -3x - 2x - 2x = 9 - 1 + 9 + 14 - 3
=> -7x = 28
=> x = 28 : (-7)
=> x = -4
Vậy ...
=>(-3x+3)-(2x+14)-9=2x+9-1
=>(-3x-2x)-(14-3+9)=2x+8
=>-5x-20=2x+8
=>2x+5x=-8-20
=>-7x=-28
=>x=-28/7
1)
a)2x+26-84+7x=-130
(2x+7x)+(-84+26)=-130
9x-58=-130
9x=-130+58
9x=-72
x=-72/9
x=-8
Vậy x=-8
b)-6x+12+12-8x=-130
(-6x-8x)+(12+12)=-130
-14x+24=-130
-14x=-130-24
-14x=-154
x=-154/(-14)
x=11
Vậy x=11
c)4x-26-8=3x-69
4x-3x=-69+26+8
x=-35
Vậy x=-35
2x^2 - 3 = 29
=>2x^2 = 29 +3 = 32
=> x^2 = \(\frac{32}{12}\)=16
x=\(\sqrt{16}\)=4
vậy x = 4
\(a,50\%x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=1\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)
\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)
\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
a) Ta có: 2x+33=-11
nên 2x=-44
hay x=-22
b) Ta có: \(\dfrac{x}{2}=\dfrac{-49}{14}\)
nên x=-7
c) Ta có: \(\dfrac{5}{6}x+\dfrac{10}{3}=\dfrac{7}{2}\)
nên \(\dfrac{5}{6}x=\dfrac{7}{2}-\dfrac{10}{3}=\dfrac{1}{6}\)
hay \(x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\)
a) 2x-12=18 x=3
2x - 12 = 18................................
2x = 30 ........................................
x = 15.....................................