Phân tích thành nhân tử:
(3-x)^4+(4-x)^4-(7-2x)^4
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\(x^4-2x^3+2x-1=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(x^3-x^2-x+1\right)=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]=\left(x-1\right)^2\left(x^2-1\right)=\left(x-1\right)^3\left(x+1\right)\)
\(x^4-2x^3+2x-1\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
x^11+x^4+1
=x^11-x^2+x^4-x+x^2+x+1
=x^2(x^9-1)+x(x^3-1)+(x^2+x+1)
=x^2[(x^3-1)(x^6+x^3+1)]+x(x-1)(x^2+x+1)+(x^2+x+1)
=x^2(x-1)(x^2+x+1)(x^6+x^3+1)+x(x-1)(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)[x^2(x-1)(x^6+x^3+1)+x(x-1)+1]
=(x^2+x+1)(x^9-x^8+x^6-x^5+x^3-x+1)
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
x^4+x^3+2x^2+x+1
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
=(x^2+1)(x^2+x+1)
\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=x^3\left(x-1\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2.\left(x^2+x+1\right)\)
x4 - x3 - x + 1
= (x4 - x3) - (x - 1)
= x3(x - 1) - (x - 1)
= (x3 - 1)(x - 1)
(3-x)4+(4-x)4-(7-2x)4
Đặt 3-x=a, 4-x=b=> 7-2x=a+b
=> (3-x)4+(4-x)4-(7-2x)4
=a4+b4-(a+b)4
=-(4a3b+6a2b2+4ab3)
=-2ab(a2+3ab+b2)
Thay a,b vào là xong nhé
Chúc
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