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\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2.x^2+1^2\right]-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+1-x\right)\left(x^2+1+x\right)\)
Tham khảo nhé~
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Bài 1 : hđt bạn tự làm nhé
Bài 2 :
\(\left(x-1\right)\left(x^2+x+1\right)-\left(x-4\right)^2x\)
\(=x^3-1-x\left(x^2-8x+16\right)=x^3-1-x^3+8x^2-16x\)
\(=8x^2-16x-1\)
\(\left(x+7\right)\left(x^2-7x+49\right)-\left(5-x\right)\left(5+x\right)\left(x-1\right)\)
\(=x^3+343-\left(25-x^2\right)\left(x-1\right)=x^3+343-\left(25x-25-x^3+x^2\right)\)
\(=x^3+343+x^3-x^2-25x+25=2x^3-x^2-25x+368\)
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1. \(x^3+8y^3\)
\(=x^3+\left(2y\right)^3\)
\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)^2\)
\(=\left(x-y\right)^3\)
2. \(a^6-b^3\)
\(=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\left(\left(a^2\right)^2+a^2b+b^2\right)\)
\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=x^4-4x^3+6x^2-4x+1 + x^4+4x^3+6x^2+4x+1
=2x^4+12x^2+2
=2(x^4+6x^2+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt y=x2+x+1
Thay y vào biểu thức ta được
y(y+1)-12
=y2 + y - 12
= y2 - 3y + 4y -12
= y(y-3) + 4(y-3)
= (y-3)(y+4)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(x^4-1=\left(x^2+1\right)\left(x^2-1\right)=\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)
Bài làm :
\(x^4-1=\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)