Question

Bài 1: phân tích đa thức thành nhân tử

a) 3x\(^3\) - 6x\(^2\) b) x\(^2\)+ 4x +4-9y\(^2\)

c)\(x^2-5x-6\) d) \(x^7+x^2+1\)

Bài 2: Tìm X

a)\(3\left(2x-4\right)+15=-11\)

b)\(x\left(x+2\right)-3x-6=0\)

d)\(\left(x+2\right)^2-\left(x-2\right)^5=64\)

Answer 1

Bài 1:

a) \(3x^3-6x^2\)

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

b) \(x^2+4x+4-9y^2\)

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

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

c) \(x^2-5x-6\)

= \(x^2+1x-6x-6\)

= \(x\left(x+1\right)-6\left(x+1\right)\)

= \(\left(x+1\right)\left(x-6\right)\)

d) \(x^7+x^2+1\)

= \(x^7-x+x^2+x+1\)

= \(x\left(x^6-1\right)+x^2+x+1\)

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

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

= \(\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)

Bài 2:

a) \(3\left(2x-4\right)+15=-11\)

⇒ \(3\left(2x-4\right)=-11-15\)

⇒ \(3\left(2x-4\right)=-26\)

⇒ \(2x-4=-26:3\)

⇒ \(2x=\frac{-26}{3}+4\)

⇒ \(2x=\frac{-14}{3}\)

⇒ \(x=\frac{-14}{3}:2\)

⇒ \(x=\frac{-7}{3}\)

b) \(x\left(x+2\right)-3x-6=0\)

⇒ \(x\left(x+2\right)-3\left(x+2\right)=0\)

⇒ \(\left(x+2\right)\left(x-3\right)=0\)

⇒ \(x+2=0\) hoặc \(x-3=0\)

TH1: \(x+2=0\)

⇒ \(x=-2\)

TH2: \(x-3=0\)

⇒ \(x=3\)

Vậy x=-2 hoặc x=3

Answer 2

Bài 1:

a) \(3x^3-6x^2\)

\(=3x^2.\left(x-2\right)\)

b) \(x^2+4x+4-9y^2\)

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

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

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

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

c) \(x^2-5x-6\)

\(=x^2-6x+x-6\)

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

\(=x.\left(x-6\right)+\left(x-6\right)\)

\(=\left(x-6\right).\left(x+1\right)\)

d) \(x^7+x^2+1\)

\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)

\(=x^5.\left(x^2+x+1\right)-x^4.\left(x^2+x+1\right)+x^2.\left(x^2+x+1\right)-x.\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right).\left(x^5-x^4+x^2-x+1\right)\)

Chúc bạn học tốt!

Answer 3

Bài 1:

a,

\(3x^3-6x^2=3x^2\left(x-2\right)\)

b, \(x^2+4x+4-9y^2=\left(x+2\right)^2-\left(3y\right)^2=\left(x-3y+2\right)\left(x+3y+2\right)\)

c,\(x^2-5x-6=\left(x^2+x\right)-\left(6x+6\right)=x\left(x+1\right)-6\left(x+1\right)=\left(x+1\right)\left(x-6\right)\)d,

\(x^2-5x-6=\left(x^2+x\right)-\left(6x+6\right)=x\left(x+1\right)-6\left(x+1\right)=\left(x+1\right)\left(x-6\right)\)

Answer 4

Bài 2: d, \(\left(x+2\right)^2-x\left(x-2\right)^5=64\)

\(\Leftrightarrow\left(x-2\right)^2+8\left(x-2\right)+16-\left(x-2\right)^5=64\)

\(\Leftrightarrow-\left(x-2\right)^5+\left(x-2\right)^2+8\left(x-2\right)-48=0\)

\(\Rightarrow\) Phương trình vô nghiệp.