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\(x^2\left(x+1\right)^2+x^2+\left(x+1\right)^2\)
\(=\left(x^2+1\right)\left(x+1\right)^2+x^2+1-1\)
\(=\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]-1\)
\(=\left(x^2+1\right)\left(x^2+1+2x+1\right)-1\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2+1-1\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)
\(=\left(x^2+1+x\right)^2\)
= (x2 + 1)(x+1)2 + x2 + 1 -1 = (x2 + 1) [(x+1)2 + 1] - 1 = (x2 + 1) (x2 + 1 + 2x + 1) - 1 = (x2 + 1)2 + 2x(x2 + 1) + x2 + 1 - 1
= (x2 + 1)2 + 2x(x2 + 1) + x2 = (x2 + 1 + x)2
Ta có:\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2=3x^4+3x^2+3-x^4-x^2-1-2x^3-2x-2x^2\)
\(=2x^4-2x^3-2x+2=2x^3\left(x-1\right)-2\left(x-1\right)=2\left(x^3-1\right)\left(x-1\right)\)
\(=2\left(x-1\right)^2\left(x^2+x+1\right)\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^4+x^2+1\right)-\left(x^4+x^2+1+2x^3+2x^2+2x\right)\)
\(=\left(x^4+x^2+1\right)\left(3-2x^3-2x^2-2x\right)\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x+x^2+x+1\right)\)
\(=3\left(x^2+x+1\right)\left(x^2+2x+1\right)\)
\(=3\left(x^2+x+1\right)\left(x+1\right)^2\)
\(\left(x+1\right)^2+\left(x^2+x+1\right)^2\)
\(\text{ Phân tích thành nhân tử}\)
\(x^4+2x^3+4x^2+4x+2\)
Chúc bạn học tốt !
(x+1)2+(x2+x+1)
=x2+2x+1+x2+x+1+x2+x+1
=(x2+x2+x2)+(2x+x+x)+(1+1+1)
=3x2+4x+3
3(x4+x+1)-(x2+x+1)2
=3(x2+x+1)(x2-x+1)-(x2+x+1)2
=(x2+x+1)[3(x2-x+1)-(x2-x+1)
=(x2+x+1)(3x2-3x+3-x2+x-1)
=(x2+x+1)(2x2-2x+2)
=(x2+x+1)2(x2-x+1)
bạn vu cong thien làm sai rồi.
\(x^4+x^2+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
chứ không phải là:
\(x^4+x+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\)đâu!
Đặt x^2 + x + x = t
Ta có BT : \(t\left(t+1\right)-1^2=t^2+t-1\):)) đề lỗi j ko ?
không bn ơi.