Tìm x:
a) 2(x-2)(x+2) + 4(x-2)(x+1) + (x+2)(8-5x)+0
b) (2x+1)(5x-1) = 20x ² – 16x-1
c) 4x(2x ²-1) + 27 = (4x ²+6x+9)(2x+3)
Tìm x:
a) 2(x-2)(x+2) + 4(x-2)(x+1) + (x+2)(8-5x)+0
b) (2x+1)(5x-1) = 20x ² – 16x-1
c) 4x(2x ²-1) + 27 = (4x ²+6x+9)(2x+3)
`a) 2(x – 2)(x + 2) + 4(x – 2)(x + 1) + (x + 2)(8 – 5x) = 0`
`=> 2(x^2 – 2^2) + 4(x^2 + x – 2x – 2) + 8x – 5x^2 + 16 – 10x = 0`
`=> 2x^2 – 8 + 4x^2 + 4x – 8x – 8 + 8x – 5x^2 + 16 – 10x = 0`
`=> (2x^2 + 4x^2 – 5x^2) + (4x – 8x + 8x – 10x) + (-8 – 8 + 16) = 0`
`=> x^2 – 6x = 0`
`=> x(x – 6) = 0`
`=>` \(\left[ \begin{array}{l}x=0\\x=6\end{array} \right.\)
Vậy `x in {0; 6}`
`b) (2x + 1)(5x – 1) = 20x^2 – 16x – 1`
`=> 10x^2 – 2x + 5x – 1 = 20x^2 – 16x – 1`
`=> 10x^2 + 3x – 1 = 20x^2 – 16x – 1`
`=> 20x^2 – 16x – 1 – 10x^2 – 3x + 1 = 0`
`=> (20x^2 – 10x^2) – (16x + 3x) + (-1 + 1) = 0`
`=> 10x^2 – 19x = 0`
`=> x(10x – 19) = 0`
`=>` \(\left[ \begin{array}{l}x=0\\10x-19=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{19}{10}\end{array} \right.\)
Vậy `x in {0; 19/10}`
`c) 4x(2x^2 – 1) + 27 = (4x^2 + 6x + 9)(2x + 3)`
`=> 8x^3 – 4x + 27 = 8x^3 + 12x^2 + 12x^2 + 18x + 18x + 27`
`=> 8x^3 + 12x^2 + 12x^2 + 18x + 18x + 27 – 8x^3 + 4x – 27 = 0`
`=> (8x^3 – 8x^3) + (12x^2 + 12x^2) + (18x + 18x + 4x) + (27 – 27) = 0`
`=> 24x^2 + 40x = 0`
`=> x(24x + 40) = 0`
`=>` \(\left[ \begin{array}{l}x=0\\24x+40=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-5}{3}\end{array} \right.\)
Vậy `x in {0; (-5)/3}`
`a)2(x-2)(x+2)+4(x-2)(x+1)+(x+2)(8-5x)=0`
`⇔2(x²-4)+(4x-8)(x+1)+8x-5x²+16-10x=0`
`⇔2x²-8+4x²+4x-8x-8+8x-5x²+16-10x=0`
`⇔(2x²+4x²-5x²)+(4x-8x+8x-10x)+(-8-8+16)=0`
`⇔x²-6x=0`
`⇔x(x-6)=0`
`(1)x=0`
`(2)x-6=0⇔x=6`
Vậy `x=0` hoặc `x=6`
`b)(2x+1)(5x-1)=20x²-16x-1`
`⇔10x²-2x+5x-1=20x²-16x-1`
`⇔10x²-2x+5x-20x²+16x=-1+1`
`⇔(10x²-20x²)+(-2x+5x+16x)=0`
`⇔-10x²+19x=0`
`⇔x(-10x+19)=0`
`(1)x=0`
`(2)-10x+19=0⇔x=19/10`
Vậy `x=0` hoặc `x=19/10`
`c)4x(2x²-1)+27=(4x²+6x+9)(2x+3)`
`⇔8x³-4x+27=8x³+12x²+18x+12x²+18x+27`
`⇔8x³-4x-8x³-12x²-18x-12x²-18x=27-27`
`⇔(8x³-8x³)+(-12x²-12x²)+(-4x-18x-18x)=0`
`⇔-24x²-40x=0`
`⇔-8x(3x+5)=0`
`(1)-8x=0⇔x=0`
`(2)3x+5=0⇔x=(-5)/3`
Vậy `x=0` hoặc `x=(-5)/3`