a) (x + 2)(3x + 1)
= x(3x + 1) + 2(3x + 1)
= 3x2 + x + 6x + 2
= 3x2 + 7x + 2
b) (5x + 2)(x - 3)
= 5x(x - 3) +2(x - 3)
= 5x2 - 15x + 2x - 6
= 5x2 - 13x - 6
c) (2x - 3)(2x + 3)
= 2x(2x + 3) - 3(2x + 3)
= 4x2 + 6x - 6x - 9
= 4x2 - 9
d) (2x - 1)(3x - 2)
= 2x(3x - 2) - 1(3x - 2)
= 6x2 - 4x - 3x + 2
= 6x2 - 7x + 2
e) (2x + 5)(x - 3) - 2x2 + x
= 2x(x - 3) + 5(x - 3) - (2x2 + x)
= 2x2 - 6x + 5x - 15 - 2x2 - x
= -2x - 15
f) (2x + 1)(3x - 2) - 62
= 2x(3x - 2) + 1(3x - 2)
= 6x2 - 4x + 3x - 2
= 6x2 -x - 2
g) (2x - 1)(2x - 1)(2x - 1)
= (2x - 1)3
h) (x + y - z)(x + y + z)
= x(x + y + z) + y(x + y + z) - x(x + y + z)
= x2 + xy + xz + xy + y2 + yz - x2 - xy - xz
= y2 + xy + xz + yz
Chuc ban hoc tot
a) Ta có: \(\left(x+2\right)\left(3x+1\right)\)
\(=3x^2+x+6x+2\)
\(=3x^2+7x+2\)
b) Ta có: \(\left(5x+2\right)\left(x-3\right)\)
\(=5x^2-15x+2x-6\)
\(=5x^2-13x-6\)
c) Ta có: \(\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+6x-6x-9\)
\(=4x^2-9\)
d) Ta có: \(\left(2x-1\right)\left(3x-2\right)\)
\(=6x^2-4x-3x+2\)
\(=6x^2-7x+2\)
e) Ta có: \(\left(2x+5\right)\left(x-3\right)-2x^2+x\)
\(=2x^2-6x+5x-15-2x^2+x\)
=-15
f) Ta có: \(\left(2x+1\right)\left(3x-2\right)-6x^2\)
\(=6x^2-4x+3x-2-6x^2\)
=-x-2
g) Ta có: \(\left(2x-1\right)\left(2x-1\right)\left(2x-1\right)\)
\(=\left(2x-1\right)^3\)
\(=8x^3-12x^2+6x-1\)
h) Ta có: \(\left(x+y-z\right)\left(x+y+z\right)\)
\(=\left(x+y\right)^2-z^2\)
\(=x^2+2xy+y^2-z^2\)
