tìm x bt : x(x-2)(x+2)-(x+2)(x^2-2x+4)=4
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HV
1
12 tháng 12 2023
a: \(2x\left(x-1\right)-x\left(2x-5\right)=9\)
=>\(2x^2-2x-2x^2+5x=9\)
=>3x=9
=>\(x=\dfrac{9}{3}=3\)
b: \(\left(3x-2\right)^2-5\left(x-1\right)\left(x+2\right)=\left(2x-3\right)^2\)
=>\(9x^2-12x+4-5\left(x^2+x-2\right)=4x^2-12x+9\)
=>\(9x^2-12x+4-5x^2-5x+10=4x^2-12x+9\)
=>\(4x^2-17x+14=4x^2-12x+9\)
=>\(-17x+14=-12x+9\)
=>\(-5x=-5\)
=>x=1
BD
1
H9
HT.Phong (9A5)
CTVHS
2 tháng 7 2023
a) \(\left(x+3\right)^2-x\left(x-1\right)=2\)
\(\Leftrightarrow x^2+6x+9-x^2+x=2\)
\(\Leftrightarrow7x+9=2\)
\(\Leftrightarrow7x=2-9\)
\(\Leftrightarrow7x=-7\)
\(\Leftrightarrow x=\dfrac{-7}{7}=-1\)
b) \(\left(2x+3\right)^2-\left(x+1\right)\left(4x-3\right)=-1\)
\(\Leftrightarrow4x^2+12x+9-\left(4x^2-3x+4x-3\right)=-1\)
\(\Leftrightarrow4x^2+12x+9-4x^2+3x-4x+3=-1\)
\(\Leftrightarrow11x+12=-1\)
\(\Leftrightarrow11x=-13\)
\(\Leftrightarrow x=\dfrac{-13}{11}\)
23 tháng 3 2020
\(A=\frac{2x^2+4x}{x^3-4x}+\frac{x^2-4}{x^2+2x}+\frac{2}{2-x}\left(x\ne0;x\ne\pm2\right)\)
\(A=\frac{2x^2+4x}{x\left(x^2-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}-\frac{2}{x-2}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}+\frac{x^3-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}-\frac{2x^2+4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x^2+4x+x^3-2x^2-4x+8-2x^2-4x}{x\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{-2x^2-4x+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x\left(x+2\right)+8}{x\left(x-2\right)\left(x+2\right)}=\frac{-2x+8}{x\left(x-2\right)}\)
Vậy \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
b) \(A=\frac{-2x+8}{x\left(x-2\right)}\left(x\ne0;x\ne\pm2\right)\)
Ta có: x=4 (tmđk) thay vào A ta có:
\(A=\frac{-2\cdot4+8}{4\left(4-2\right)}=\frac{-8+8}{4\cdot2}=\frac{0}{8}=0\)
Vậy A=0 với x=4
H
0
AL
4
21 tháng 8 2018
a, ( 8x - 3 ) ( 3x + 2 ) - ( 4x + 7 ) ( x + 4 ) = ( 2x + 1 ) ( 5x - 1 )
( 24x2 + 16x - 9x - 6 ) - ( 4x2 - 16x - 7x + 28 ) = 10x2 - 2x + 5x -1
24x2 + 16x - 9x - 6 -4x2 - 16x - 7x - 10x2 + 2x - 5x = 6 + 28 - 1
10x2 -19x = 33
10x2 - 19x -33 = 0 \(\Leftrightarrow\)10x( x+ 3 ) + 11 ( x- 3 ) = 0
=> ( x- 3 ) ( 10x + 11 ) = 0\(\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{-11}{10}\end{cases}}\)
b, 4( x - 1 ) ( x + 5 ) - ( x + 2 ) ( x + 5 ) = 3( x - 1 ) ( x + 2 )
4( x2 - 5x - x + 5 ) - ( x2 + 5x + 2x + 10 ) = 3( x2 + 2x - x - 2 )
4x2 - 20x - 4x + 20 - x2 - 5x - 2x - 10 = 3x2 + 6x - 3x - 6
( 4x2 - x2 ) + ( -20x - 4x - 5x - 2x ) + 20 - 10 = 3x2 + ( 6x - 3x ) - 6
3x2 - 31x - 3x2 - 3x = -6-10
-34x = -16
x = \(\frac{8}{17}\)
\(x\left(x-2\right)\left(x+2\right)-\left(x+2\right)\left(x^2-2x+4\right)=4\)
\(\Rightarrow x\left(x^2-4\right)-\left(x^3+8\right)=4\)
\(\Rightarrow x^3-4x-x^3-8=4\)
\(\Rightarrow4x=-12\Rightarrow x=-3\)
\(\Leftrightarrow x^3-4x-x^3-8=4\)
\(\Leftrightarrow-4x=12\)
hay x=-3