Phân tích đa thức thành nhân tử, giúp mình vớiiii
x8 + x +1
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1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
\(x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)
\(=\left(x+1\right)\left(x+2\right)\left[x\left(x+3\right)\right]+1\)
\(=\left(x^2+x+2x+2\right)\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+2\right)\left(x^2+3x\right)+1\)
gọi \(\left(x^2+3x\right)=a\)
\(\Rightarrow\left(t+2\right)t+1\)
\(=t^2+2t+1=\left(t+1\right)^2\)
\(\Rightarrow=\left(x^2+3x+1\right)^2\)
\(\Rightarrow x\cdot\left(x+1\right)\cdot\left(x+2\right)\cdot\left(x+3\right)+1\)\(=\left(x^2+3x+1\right)^2\)
\(\left(4x^2-4x+1\right)-\left(x-1\right)^2\)
\(=\left(2x-1\right)^2-\left(x-1\right)^2\)
\(=\left(2x-1-x+1\right)\left(2x-1+x-1\right)\)
\(=x\left(3x-2\right)\)
Ta có :
\(x^4-3x^2+1\)
\(=\left(x^4-2x^2+1\right)-x^2\)
\(=\left(x^2-1\right)^2-x^2\)
\(=\left(x^2-1-x\right)\left(x^2-1+x\right)\)
\(=x^3+2x^2-8x=x\left(x^2+2x-8\right)\\ =x\left(x^2-2x+4x-8\right)\\ =x\left(x-2\right)\left(x+4\right)\)
x11 + x10 + 1
= ( x11 - x9 + x8 - x6 + x5 - x3 + x2 ) + ( x10 - x8 + x7 - x5 + x4 - x2 + x ) + ( x9 - x7 + x6 - x4 + x3 - x + 1 )
= x2 ( x9 - x7 + x6 - x4 + x3 - x + 1 ) + x( ( x9 - x7 + x6 - x4 + x3 - x + 1 ) + 1( x9 - x7 + x6 - x4 + x3 - x + 1 )
= ( x2 + x + 1 ) ( x9 - x7 + x6 - x4 + x3 - x + 1 )
Chỗ nào không hiểu thì ib nhé :)
x11+x10+1
=x11+x10+x9-x9-x8-x7+x8+x7+x6-x6-x5-x4+x5+x4+x3-x3-x2-x+x2+x+1
=(x11+x10+x9)-(x9+x8+x7)+(x8+x7+x6)-(x6+x5+x4)+(x5+x4+x3)-(x3+x2+x)+(x2+x+1)
=x9(x2+x+1)-x7(x2+x+1)+x6(x2+x+1)-x4(x2+x+1)+x3(x2+x+1)-x(x2+x+1)+(x2+x+1)
=(x2+x+1)(x9-x7+x6-x4+x3-x+1)
A=x14+x7+1
=(x14+x13+x12)-(x13+x12+x11)+(x11+x10+x9)-(x10+x9+x8)+(x8+x7+x6)-(x6+x5+x4)+(x5+x4+x3)-(x3+x2+x)+(x2+x+1)
Đặt B=x2+x+1
=>A=x12B-x11B+x9B-x8B+x6B-x4B+x3B-xB+B
=>A=B(x12-x11+x9-x8+x6-x4+x3-x+1)
Thay B=x2+x+1 vào A là xong
Lời giải :
\(x^8+x+1\)
\(=x^8-x^5+x^5-x^2+x^2+x+1\)
\(=x^5\left(x^3-1\right)+x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x-1\right)\left(x^2+x+1\right)+x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
Mơn bạn nhiều nhé =))