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31 tháng 10 2022

Bài 2:

a: \(=\left(4x^2+6x+2x+3\right)\left(x^2+2x+1\right)-18\)

\(=\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18\)

Đặt x^2+2x=a

\(A=\left(4a+3\right)\left(a+1\right)-18\)

\(=4a^2+7a-15\)

\(=4a^2+12a-5a-15=\left(a+3\right)\left(4a-5\right)\)

\(=\left(x^2+2x+3\right)\left(4x^2+8x-5\right)\)

\(=\left(x^2+2x+3\right)\left(4x^2+10x-2x-5\right)\)

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

b: \(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

28 tháng 10 2018

 \(A=\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)-18\)

\(=\frac{1}{4}\left[\left(2x+1\right)\left(x+1\right)^2.4\left(2x+3\right)\right]-72\)

\(=\frac{1}{4}\left[\left(2x+1\right)\left(2x+3\right)\left(2x+2\right)^2\right]-72\)

\(=\frac{1}{4}\left[\left(4x^2+8x+3\right)\left(4x^2+8x+4\right)-72\right]\)

Đặt: \(4x^2+8x+3=t\)

Ta có:  \(A=\frac{1}{4}\left[t^2+t-72\right]\)

\(=\frac{1}{4}\left[\left(t+9\right)\left(t-8\right)\right]\)

\(=\frac{1}{4}\left[\left(4x^2+8x+12\right)\left(4x^2+8x-5\right)\right]\)

\(=\left(x^2+2x+3\right)\left[4x^2+8x-5\right]\)

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

 \(B=\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)

\(=\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x+2=a\)

Khi đó: \(B=a\left(a-3\right)-4\)

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

\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)

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

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

\(=x^4-2x^3+6x^2-8x+8\)

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

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

       \(3x^4-5x^3-18x^2-3x+5\)

\(=3x^4+x^3-x^2-6x^3-2x^2+2x-15x^2-5x+5\)

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

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

Bài này thật sự khó và hay đấy.

15 tháng 9 2021

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

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

15 tháng 9 2021

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

4 tháng 11 2016

b)(x2+x+1)(x2+x+2)-12

Đặt t=x2+x+1

t(t+1)-12=t2+t-12

=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)

=(x2+x-2)(x2+x+5)

=(x-1)(x+2)(x2+x+5)

c)(x2+8x+7)(x2+8x+15)+15

Đặt t=x2+8x+7 

t(t+8)+15=t2+8t+15

=(t+3)(t+5)

=(x2+8x+7+3)(x2+8x+7+15)

=(x2+8x+10)(x2+8x+22)

d)(x+2)(x+3)(x+4)(x+5)-24

=(x2+7x+10)(x2+7x+12)-24

Đặt t=x2+7x+10

t(t+2)-24=(t-4)(t+6)

=(x2+7x+10-4)(x2+7x+10+6)

=(x2+7x+6)(x2+7x+16)

=(x+1)(x+6)(x2+7x+16)

4 tháng 11 2016

a/ Đặt x2 + 4x + 8 = a

Thì đa thức ban đầu thành

a2 + 3ax + 2x= (a2 + 2ax + x2) + (ax + x2)

= (a + x)2 + x(a + x) = (a + x)(a + 2x)

12 tháng 9 2017

đặt \(x^2+4x+8=a\)

=> \(A=a^2+3ax+2x^2=a^2+ax+2ax+2x^2=a\left(a+x\right)+2x\left(a+x\right)\)

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

b) ta có 

\(B=\left(x+1\right)\left(x+7\right)\left(x+3\right)\left(x+5\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

đặt \(x^2+8x+11=a\)

=> \(B=\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a-1\right)\left(a+1\right)\)

         \(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x^2+6x+2x+12\right)\)

         \(=\left(x^2+8x+10\right)\left[x\left(x+6\right)+2\left(x+6\right)\right]=\left(x^2+8x+10\right)\left(x+6\right)\left(x+2\right)\)

12 tháng 9 2017

khó thế

2 tháng 11 2016

a)x4+2x3+5x2+4x-12

=(x4+2x3+x2)+(4x2+4x)-12

=(x2+x)2+4(x2+x)-12

Đặt t=x2+x

=t2+4t-12=(t-2)(t+6)

=(x2+x-2)(x2+x+6)

=(x-1)(x+2)(x2+x+6)

b)(x+1)(x+2)(x+3)(x+4)+1

=(x2+5x+4)(x2+5x+6)+1

Đặt x2+5x+4=t

t(t+2)+1=t2+2t+1

=(t+1)2=(x2+5x+4+1)2

=(x2+5x+5)2

c)(x+1)(x+3)(x+5)(x+7)+15

=(x2+8x+7)(x2+8x+15)+15

Đặt t=x2+8x+7

t(t+8)+15=(t+3)(t+5)

=(x2+8x+7+3)(x2+8x+7+5)

=(x2+8x+10)(x+2)(x+6)

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

=(x2+5x+4)(x2+5x+6)-24

Đặt t=x2+5x+4 

t(t+2)-24=(t-4)(t+6)

=(x2+5x+4-4)(x2+5x+4+6)

=x(x+5)(x2+5x+10)

11 tháng 9 2018

Đặt x^2+2x=t =>3t^2-2t-1=3t^2-3t+t-1=3t(t-1)+(t-1)=(t-1)(3t+1)

=>(x^2+2x-1)(3x^2+6x+1)

16 tháng 10 2020

(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2

= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2

= (x - 5 + 2x - 1)2 - (2x - 6)2

= (3x - 6)2 - (2x - 6)2

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

16 tháng 10 2020

( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2

= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2

= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2

= ( 3x - 6 )2 - ( 2x - 6 )2

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )