Question

Answer 1

Bài 4: 

Ta có: a+b=11

<=>a=11-b (*)

Thay * vào pt tích ta có:

(11-b).b=30

<=> 11b-b²=30

<=>-b²+11b-30=0

<=>\(\left\{{}\begin{matrix}b=5\\b=6\end{matrix}\right.\)

Vậy => b=5, a=6 hoặc b=6, a=5

a)Thay a=5, b =6 vào bt ta có:

5²+6²

=25+36

=61

b)Thay a=5, b=6 vào bt ta có:

(5-6)¹⁰⁰

=(-1)¹⁰⁰

=1

Answer 2

Bài 2: 

a) Ta có: \(\left(2x+1\right)^2-9\)

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

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

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

b) Ta có: \(\left(x-3\right)^2-\left(2x+3\right)^2\)

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

\(=3x\left(-x-6\right)\)

c) Ta có: \(x^3-x^2-12x\)

\(=x\left(x^2-x-12\right)\)

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

d) Ta có: \(x^3-4x^2y+4xy^2-9x\)

\(=x\left(x^2-4xy+4y^2-9\right)\)

\(=x\left(x-2y-3\right)\left(x-2y+3\right)\)