Question

1.Phân tích đa thức thành nhân tử:

a) 9x²+6x-8

b) x²-7xy+10y²

c) a⁴+4b⁴

d) (x+1)(x+2)(x+3)(x+4)-24

e) x²+19x+84

2.Tìm x biết:

a) x²-25+3(x-5)²=0

b) x³-3x²+3x=1

Answer 1

Bài 2: 

a: \(x^2-25+3\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\left(x-5\right)\left(3x-15\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+5+3x-15\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(4x-10\right)=0\)

hay \(x\in\left\{5;\dfrac{5}{2}\right\}\)

b: \(x^3-3x^2+3x=1\)

\(\Leftrightarrow x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

=>x-1=0

hay x=1