\(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

\(=\left(2x^2+34x+120\right)\left(2x^2+32x+60\right)-3x^2\)

\(=\left(2x^2+33x+120\right)^2-x^2-3x^2\)

\(=\left(2x^2+33x+120-2x\right)\left(2x^2+33x+120+2x\right)\)

\(=\left(2x+15\right)\left(x+8\right)\left(2x^2+35x+120\right)\)

4( x+5) ( x+6) (x+10) ( x+12) -3x²

=4(x+5)(x+12)(x+6)(x+10)-3x²

=4.(x²+17x+60)(x²+16x+60)-3x²

Đặt t=x²+16x+60 ta được:

4.(t+x).t-3x²

=4t²+4tx-3x²

=4t²-2tx+6tx-3x²

=2t.(2t-x)+3x.(2t-x)

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

thay t=x²+16x+60 ta được:

[2.(x²+16x+ 60)-x][2.(x²+16x+60)+3x]

=(2x²+32x+120-x)(2x²+32x+120+3x)

=(2x²+31x+120)(2x²+35x+120)

=(2x²+16x+15x+120)(2x²+35x+120)

=[2x.(x+8)+15.(x+8)](2x²+35x+120)

=(x+8)(2x+15)(2x²+35x+120)

4( x+5) ( x+6) (x+10) ( x+12) -3x 2

=4(x+5)(x+12)(x+6)(x+10)-3x 2

=4.(x 2+17x+60)(x 2+16x+60)-3x 2

Đặt t=x 2+16x+60 ta được: 4.(t+x).t-3x 2

=4t 2+4tx-3x 2

=4t 2 -2tx+6tx-3x 2 

=2t.(2t-x)+3x.(2t-x)

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

thay t=x 2+16x+60 ta được: [2.(x 2+16x+ 60)-x][2.(x 2+16x+60)+3x]

=(2x 2+32x+120-x)(2x 2+32x+120+3x)

=(2x 2+31x+120)(2x 2+35x+120)

=(2x 2+16x+15x+120)(2x 2+35x+120)

=[2x.(x+8)+15.(x+8)](2x 2+35x+120)

=(x+8)(2x+15)(2x 2+35x+120)

\(4.\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)-3x^2\)

\(=4.\left[\left(x+5\right)\left(x+12\right)\right].\left[\left(x+6\right)\left(x+10\right)\right]-3x^2\)

\(=4.\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

Đặt \(a=x^2+16x+60\) ta có :

\(4a.\left(a+x\right)-3x^2=4a^2+4ax+x^2-4x^2=\left(2a+x\right)^2-\left(2x\right)^2\)

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

Thay a , ta có ;

\(\left(2a-x\right)\left(2a+3x\right)=\left[2.\left(x^2+16x+60\right)-x\right].\left[2.\left(x^2+16x+60\right)+3x\right]\)

\(=\left(2x^2+32x+120-x\right)\left(2x^2+32x+120+3x\right)\)

\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

\(=\left(2x^2+16x+15x+120\right)\left(2x^2+35x+120\right)\)

\(=\left[2x.\left(x+8\right)+15.\left(x+8\right)\right]\left(2x^2+35x+120\right)\)

\(=\left(x+8\right)\left(2x+15\right)\left(2x^2+35x+120\right)\)

Đây là một dạng phân tích thừa số nguyên tố khá quen, cô sẽ hướng dẫn e nhé :) Ta cần ghép các hạng tử để xuất hiện các thành phần chứa biến giống nhau.

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

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

Đặt \(12x^2+11x+2=t\Rightarrow A=t\left(t-3\right)-4=t^2-3t-4=\left(t-4\right)\left(t+1\right)\)

Quay lại biến x ta có: \(A=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

Câu sau tương tự nhé :)

a) đề thế này\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)

Đặt \(x^2+7x+11=t\)vào (1) ta được:

\(\left(t-1\right)\left(t+1\right)-24\)

\(=t^2-1-24\)

\(=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

b) Phân tích sẵn rồi còn phân tích gì nưa=))

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)( Làm đề theo Lê Tài Bảo Châu )

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left[\left(x^2+7x+11\right)-1\right]\left[\left(x^2+7x+11\right)+1\right]-24\)

\(=\left(x^2+7x+11\right)^2-1-24\)

\(=\left(x^2+7x+11\right)^2-25\)

\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)