Rút gọn biểu thức
a) (x + 2) . (x - 2) - (x - 3) . (x + 1).
b) (2x + 1)2 + (3x - 1)2 + 2 . (2x + 1) . (3x - 1)
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a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)
=3x-2-2x^2+2x-5x+5
=-2x^2+3
b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)
c: =x^3-3x^2+3x-1-x^3-1+9x^2-1
=6x^2+3x-3
\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)
\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)
\(=-2x^2+3\)
\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)
\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)
\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)
\(=\left(2x+1\right)\left(4x-5\right)\)
\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)
\(=x^3-3x^2+3x-1-x^3-1-\left(3x-9x^2+1-3x\right)\)
\(=-3x^2+3x-2-3x+9x^2-1+3x\)
\(=6x^2+3x-3\)
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6
=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = 9 => x = 0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0
=> 3(6x2 - 5x +1) - (18x2 - 29x + 3) = 0
=> (18x2 -15x + 1) -(18x2 - 29x +3) = 0
=> 18x2 - 15x +1 -18x2 + 29x - 3 = 0
=> 14x = 0 => x = 0
a)(x+2)(x+3)-(x-2)(x+5)=6
x(x+3)+2(x+3)-x(x+5)+2(x+5)=6
x2+3x+2x+6-x2-5x+2x+10=6
(x2-x2)+(3x+2x-5x+2x)+(10+6)=6
2x+16=6
2x=6-16
2x=-10
x=-10/2
x=-5. Vậy x=-5
b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6
6x2+27x+4x+18-6x2-x-12x-2=7
(6x2-6x2)+(27x+4x-x-12x)+(18-2)=7
18x+16=7
18x=7-16
x=-9/18=-1/2. Vậy x=-1/2
c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0
(9x-3)(2x-1)-(2x-3)(9x-1)=0
9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0
18x2-9x-6x+3-18x2+2x+27x-3=0
(18x2-18x2)+(27x+2x-6x-9x)+(3-3)=0
14x=0
x=0/14
x=0. Vậy x=0
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6 => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = -9 => x = -0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0
=> 3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
=> 18x2 - 15x + 3 - 18x2 + 29x -3 = 0
=> 14x = 0 => x = 0.
\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)
\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)
a, (x + 2) . (x - 2) - (x - 3) . (x + 1).
= x2- 4 -( x2 + x -3x -3)
= x2 - 4 -x2 -x+ 3x + 3
= 2x-1
b.(2x + 1)2 + (3x - 1)2 + 2 . (2x + 1) . (3x - 1)
=( 2x +1 + 3x-1)2
=(5x)2
= 25x2