Tìm x:
a)3.(x+2)-6.(x-5)=2.(5-2x)
b)(2x).(-4x)+28=100
c)4x\(^3\)=4x
d)5x.(-x)\(^2\)+1=6
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a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
Ta co´ :
a, 3 . ( x + 2 ) - 6 . ( x - 5 ) = 2 . ( 5 - 2x )
. 3x + 6 - 6x + 30 = 10 - 4x
3x - 6x + 4x = 10 - 6 - 30
x = - 26
b, ( - 2x ) . ( - 4x ) + 28 = 100
8x + 28 = 100
8x = 100 - 28
8x = 72
x = 72 : 8
x = 9
c, 5x . ( - x )2 + 1 = 6
5x . ( - x ) . ( - x ) + 1 = 6
5x . [ ( - x ) . ( - x ) ] + 1 = 6
5x . 2x + 1 = 6
5x . 2x = 6 - 1
5x . 2x = 5
Kẻ bảng . Co´ 4 trường hợp laˋ 5x = 1 ; 2x = 5
5x = 5. ; 2x = 1
5x = - 1 ; 2x = - 5
5x = - 5 ; 2x = - 1
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)
\(< =>20x-5x^2+5x^2-12-x-6=0\)
\(< =>19x-18=0\)
\(< =>x=\dfrac{18}{19}\)
\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)
\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)
\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)
a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)
b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)
a) 3x(4x - 3) - 2x(5 - 6x) = 0
=> 6x2 - 9x - 10x + 12x2 = 0
=> 18x2 - 19x = 0
=> x(18x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\18x-19=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{19}{18}\end{cases}}\)
b) 5(2x - 3) + 4x(x - 2) + 2x(3 - 2x) = 0
=> 10x - 15 + 4x2 - 8x + 6x - 4x2 = 0
=> 8x - 15 = 0
=> 8x = 15
=> x = 15 : 8 = 15/8
c) 3x(2 - x) + 2x(x - 1) = 5x(x + 3)
=> 6x - 3x2 + 2x2 - 2x = 5x2 + 15x
=> 4x - x2 - 5x2 - 15x = 0
=> -6x2 - 11x = 0
=> -x(6x - 11) = 0
=> \(\orbr{\begin{cases}-x=0\\6x-11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{11}{6}\end{cases}}\)
a) \(3x\left(4x-3\right)-2x\left(5-6x\right)=0\)
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow-19x=0\Leftrightarrow x=0\)
b) \(5\left(2x-3\right)+4x\left(x-2\right)+2x\left(3-2x\right)=0\)
\(\Leftrightarrow10x-15+4x^2-8x+6x-4x^2=0\)
\(\Leftrightarrow8x-15=0\Leftrightarrow x=\frac{15}{8}\)
b, (2x) . (-4x) + 28 = 100
2x . -4x = 100 - 28
-8x2 = 72
x2 = 72 : -8
x2 = -9
=> \(x\in\varphi\)
Chắc z á :v ~
a, 3( x + 2) - 6( x - 5 ) = 2( 5 - 2x )
3x + 6 - 6x + 30 - 10 + 4x = 0
<=> x( 3 - 6 + 4 ) + 6 + 30 - 10 = 0
=> x + 6 + 30 - 10 = 0
=> x= -26