Giải hộ mk vs ạ
Tìm X biết
2x+12=3(x-7)
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a) \(A=x^3+2x^2+7x-4-x-x^3-2x^2+1\)
\(A=\left(x^3-x^3\right)+\left(2x^2-2x^2\right)+\left(7x-x\right)+\left(-4+1\right)\)
\(A=6x-3\)
b) Thay x = (-5)
\(\Rightarrow A=6.\left(-5\right)-3\)
\(\Rightarrow A=-30-3\)
\(\Rightarrow A=-33\)
c) \(A=6x-3\)
\(10=6x-3\)
\(13=6x\)
\(x=\frac{13}{6}\)
a: =>\(2x+7\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
=>\(x\in\left\{-3;-4;-\dfrac{5}{2};-\dfrac{9}{2};-2;-5;-\dfrac{3}{2};-\dfrac{11}{2};-\dfrac{1}{2};-\dfrac{13}{2};\dfrac{5}{2};-\dfrac{19}{2}\right\}\)
b: =>x+2+5 chia hết cho x+2
=>\(x+2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-1;-3;3;-7\right\}\)
5 - ( 2x - 7 ) = 3 . ( -4) + 20
5 - ( 2x - 7 ) = ( -12 ) + 20
5- ( 2x - 7 ) = 8
2x - 7 = 5 - 8
2x - 7 = -3
2x = -3 + 7
2x = 4
x = 4 : 2
x = 2
chúc bn hok tốt
b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)
<=> 13(x + 1) - 2(5x + 3) = x + 7
<=> 13x + 13 - 10x - 6 = x + 7
<=> 3x + 7 = x + 7
<=> 3x + 7 - x = 7
<=> 2x + 7 = 7
<=> 2x = 7 - 7
<=> 2x = 0
<=> x = 0
c) 2x + 4(x - 2) = 5
<=> 2x + 4x - 8 = 5
<=> 6x - 8 = 5
<=> 6x = 5 + 8
<=> 6x = 13
<=> x = 13/6
\(-7-5-x=12+\left(-5+8\right)\)
\(\Leftrightarrow-12-x=12+3\)
\(\Leftrightarrow-12-x=16\)
\(\Leftrightarrow x=-12-16\)
\(\Leftrightarrow x=-28\)
\(\text{Vậy }x=-28\)
-7-5-x=12+(-5+8)
-7-5-x=15
5-x=-7-15
5-x=-7+(-15)
5-x=-(7+15)
5-x=-22
x=5-(-22)
x=5+22
x=27
Vậy x=27
a) 3x + 27 = 9
3x = 9 - 27
3x = -18
x = -18 : 3
x = - 6
Vậy x=-6
b) 2x + 12 = 3(x - 7 )
2x + 12 = 3x - 21
12 + 21 = 3x - 2x
33 = x
Vậy x=33
c) 2x2 - 1 = 49
2x2=49+1
2x2=50
x2=50:2
x2=25
x2=52
=> x= + 5
Vậy x=+5
d) |x + 9 | . 2 =10
|x + 9 | = 10 : 2
|x + 9 | = 5
\(\Rightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-9=-4\\x=-5-9=-14\end{cases}}\)
Vậy \(x\in\left\{-4;-14\right\}\)
x.(2x^2+5x-3)=0
x.(2x^2-x+6x-3)=0
x.(2x-1).(x+3)=0
-> x=0 hoặc x=-3 hoặc x=1/2
\(2x+12=3.\left(x-7\right)\)
=> \(2x+12=3x-21\)
=> \(2x-3x=-21-12=\left(-21\right)+\left(-12\right)=-33\)
=> \(x\left(2-3\right)=-33\)
=> \(x.\left(-1\right)=-33\)
=> \(x=\left(-33\right):\left(-1\right)=33\)
Vậy: x = 33
2x+12=3(x-7)
<=>2x+12=3x-21
<=>2x-3x=-21-12
<=>-x=-33
=>x=33
Vậy x=33