\(^2\) - 5x + 2

2. a

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Bài 1: Phân tích đa thức thành nhân tử

a) Ta có: \(8a^3-6a^2-1+3a\)

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

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

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

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

b) Ta có: \(x^3-2x^2y+xy^2-9x\)

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

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

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

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

c) Ta có: \(5x^2-45\)

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

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

d) Ta có: \(2x^3-4x^2+2x\)

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

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

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

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

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

e) Ta có: \(6x\left(3x-2\right)-12\left(2-3x\right)\)

\(=6x\left(3x-2\right)+12\left(3x-2\right)\)

\(=\left(3x-2\right)\left(6x+12\right)\)

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

f) Ta có: \(4x^2-8xy+4y^2-10\)

\(=\left(2x\right)^2-2\cdot2x\cdot2y+\left(2y\right)^2-10\)

\(=\left(2x-2y\right)^2-10\)

\(=\left(2x-2y-\sqrt{10}\right)\left(2x-2y+\sqrt{10}\right)\)

g) Ta có: \(2x^2-8x+8\)

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

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

h) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)

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

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

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

1) \(x^2+6x+8\)

\(=x^2+2x+4x+8\)

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

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

2) \(x^2-5x-14\)

\(=x^2-7x+2x-14\)

\(=x\left(x-7\right)+2\left(x-7\right)\)

\(=\left(x-7\right)\left(x+2\right)\)

3) \(2x^2+5x+3\)

\(=2x^2+2x+3x+3\)

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

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

4) \(x^2-x-12\)

\(=x^2-4x+3x-12\)

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

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

4 tháng 12 2018

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

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

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

b) \(\left(x^2+2\right)^2-\left(x+2\right)\left(x-2\right)\left(x^2+4\right)\)

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

\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)

\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)

\(=x^4+4x^2+4-x^4+16\)

\(=4x^2+20\)

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

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

\(=\left(4x^2-4xy+y^2\right)-2\left(x^2+6xy+9y^2\right)-\left(9x^2-1\right)\)

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

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

\(=\left(x^6-3x^4+3x^2-1\right)-\left(x^6-1\right)\)

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

\(=-3x^4+3x^2\)

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

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

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

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

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

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

\(=\left(-2\right)^2=4\)

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

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

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

\(=\left(x+2z\right)^2\)

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\)

\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)

\(=\left(2x+3-2x-5\right)^2\)

\(=\left(-2\right)^2=4\)

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)

\(=5x^2-\left(-2x^2-3x+1\right)-3\left(x+1\right)\)

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

\(=7x^2-4\)

5 tháng 8 2016

a) \(45+x^3-5x^2-9x\)

\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)

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

\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)

 

 

5 tháng 8 2016

\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)

24 tháng 9 2017

Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@

B1: Phân tích thành nhân tử:

a) \(6x^2+9x=3x\left(2x+3\right)\)

b) \(4x^2+8x=4x\left(x+2\right)\)

c) \(5x^2+10x=5x\left(x+2\right)\)

d) \(2x^2-8x=2x\left(x-4\right)\)

e) \(5x-15y=5\left(x-3y\right)\)

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

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

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

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

h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)

i) \(9x^2-18x+9=\left(3x-3\right)^2\)

k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)

l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)

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

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

n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)

\(=\left(x-31\right)\left(x+1\right)\)

o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)

\(=\left(2+x\right)\left(8+x\right)\)

p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

24 tháng 9 2017

Bài 1 :

a ) \(6x^2+9x=3x\left(x+3\right)\)

b ) \(4x^2+8x=4x\left(x+2\right)\)

c ) \(5x^2+10x=5x\left(x+2\right)\)

d ) \(2x^2-8x=2x\left(x-4\right)\)

e ) \(5x-15y=5\left(x-3y\right)\)

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

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

h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)

i ) \(9x^2-18x+9=\left(3x-3\right)^2\)

k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)

l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)

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

n ) \(\left(x-15\right)^2=x^2-30x+15^2\)

o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)

p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)

Bài 2 :

a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)

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

c ) \(2x+x^2-2y-2xy=......................\)

d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)

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

f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)

20 tháng 10 2020

Câu 1) xem lại đề giùm đi em.

27 tháng 10 2021

helpppppp

29 tháng 6 2017

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

2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)

3)\(-x^2+10x-25=-\left(x-5\right)^2\)

4)\(1+12x+36x^2=\left(1+6x\right)^2\)

5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)

6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)

29 tháng 6 2017

bài toán iêu cầu j z ??? bn