x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

ngu dốt

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

mik chịu

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

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

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

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

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

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

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

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

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

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

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

\(=\left(2x^2+120+33x\right)^2-4x^2\)

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

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

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

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)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)