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b) \(\left(2x^2+2x+1\right)\left(2x^2-2x-1\right)+\left(2x+1\right)^2\)
\(=4x^4-\left(2x+1\right)^2+\left(2x+1\right)^2\)
\(=4x^4\)
a) \(\left(3x^2+3x+1\right)\left(3x^2-3x+1\right)-\left(3x^2+1\right)^2\)
\(=\left(3x^2+1\right)^2-9x^4-\left(3x^2+1\right)^2\)
\(=-9x^4\)
\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2=\left(\left(3x+1\right)-\left(3x-5\right)\right)^2=6^2=36\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)=\left(x^2-2y\right).5x^2\)
a. BT= ((3x+1) - (3x-5))2=62=36
b. BT = (3x2-y-2x2-y). (3x2- y + 2x2+ y) = (x2-2y).5x2
a) (x+2)(x−2)−(x−3)(x+1)
=x2−22−(x2+x−3x−3)
=x2−4−x2−x+3x+3
=2x−12x−1
b) (2x+1)2+(3x−1)2+2(2x+1)(3x−1)(
=(2x+1)2+2.(2x+1)(3x−1)+(3x−1)2
=[(2x+1)+(3x−1)]2
= (2x+1+3x−1)2
=(5x)2=25x2
\(\left(2x+1\right)^2+\left(3x-1\right)^2+2\left(2x+1\right)\left(3x-1\right)\))
= \(\left(2x+1\right)^2+2\left(2x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
= \(\left[\left(2x+1\right)+\left(3x-1\right)\right]^2\)
= \(\left[2x+1+3x-1\right]^2\)
=\(\left(5x\right)^2\)= \(25x^2\)
a) \(4x^2\left(5x^3-2x+3\right)\)
\(=20x^5-8x^3+12x^2\)
b) \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)\)
\(=12y^5+2y^4-y^2\)
c) \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-14x^2+8x\)
d) \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+22x-55-6x^2-23x-21\)
\(=-x-76\)
1, \(4x^2\left(5x^3-2x+3\right)=20x^5-8x^3+12x^2\)
2, \(3y^2\left(4y^3+\frac{2}{3}y^2-\frac{1}{3}\right)=12y^5+2y^4-y^2\)
3, \(\left(5x^2-4x\right)\left(x-2\right)=5x^3-10x^2-4x^2+8x=5x^3-14x^2+8x\)
4, \(\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)=6x^2+33x-10x-55-\left(6x^2+14x+9x+21\right)\)
\(=6x^2+23x-55-6x^2-23x-21=-76\)
1) (2x - 1)2 - (x + 1)(3x - 2)
= 4x2 - 4x + 1 - 3x2 - x + 2
= (4x2 - 3x2) + (-4x - x) + (1 + 2)
= x2 - 5x + 3
2) (3x - 2)(-x) - (-2x)2
= -3x2 + 2x - 4x2
= -7x2 + 2x
a)\(\frac{x^3-x}{3x+3}=\frac{x.\left(x^2-1\right)}{3.\left(x+1\right)}=\frac{x.\left(x-1\right).\left(x+1\right)}{3.\left(x+1\right)}=\frac{x.\left(x+1\right)}{3}=\frac{x^2+x}{3}\)
(2x + 1)2 + (3x – 1)2 + 2(2x + 1)(3x – 1)
= (2x + 1)2 + 2.(2x + 1)(3x – 1) + (3x – 1)2
= [(2x + 1) + (3x – 1)]2
= (2x + 1 + 3x – 1)2
= (5x)2
= 25x2