Question

Bài 3 : Phân tích các đa thức sau thành nhân tử
a, (3x-1)^2 - 16
b, (5x-4)^2 - 49x^2
c, (2x+5)^2 - (x-9)^2
d, (3x+1)^2 - 4(x-2)^2
e, 9(2x+3)^2 - 4(x+1)^2
f, 4b^2c^2 - (b^2 + c^2 - a^2 )

Giúp mk vs ạ mk đang cần gấp ạ

Answer 1

Sửa lại ạ!

a) \(\left(3x-1\right)^2-16\)

\(=\left(3x-1\right)^2-4^2\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

\(=\left(3x-5\right)\left(3x+3\right)\)

b) \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4\right)^2-\left(7x\right)^2\)

\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)

\(=\left(-4-2x\right)\left(-4+12x\right)\)

c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)

\(=\left(x+14\right)\left(3x-4\right)\)

d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)

\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)

\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)

\(=\left(x+5\right)\left(5x-3\right)\)

e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)

\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)

\(=\left(4x+7\right)\left(8x+11\right)\)

P/s: Ko chắc!

Answer 2

a) \(\left(3x-1\right)^2-16\)

\(=\left(3x-1\right)^2-4^2\)

\(=\left(3x-1-4\right)\left(3x-1+4\right)\)

b) \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4\right)^2-\left(7x\right)^2\)

\(=\left(5x-4+7x\right)\left(5x-4-7x\right)\)

c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)

\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)

d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)

\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)

\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)

\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)

e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)

\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9-2x+2\right)\left(6x+9+2x-2\right)\)