Phân tích đa thức sau thành nhân tử
\(^{x^4}\)+ 64
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\(3x^2+x-4=3x^2-3x+4x-4=3x\left(x-1\right)+4\left(x-1\right)=\left(3x+4\right)\left(x-1\right)\)
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
\(\text{a) }x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^4+16x^2+64\right)-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2+4x+8\right)\left(x^2-4x+8\right)\)
\(\text{b) }4x^4+81y^4\)
\(=4x^4+36x^2y^2+81y^4-36x^2y^2\)
\(=\left(4y^4+36x^2y^2+81y^4\right)-36x^2y^2\)
\(=\left(2x^2+9y^2\right)^2-\left(6xy\right)^2\)
\(=\left(2x^2+9y^2+6xy\right)\left(2x^2+9y^2-6xy\right)\)
a. x4 + 64
= (x2)2 + 2x28 + 82 - 2x28
= (x2 + 8)2 - (4x)2
= (x2 + 8 + 4x)(x2 + 8 - 4x)
b. 4x4 + 81y4
= (2x2)2 + (9y2)2
Làm tới đây bí rồi bạn! Mà hình như làm gì có công thức a2 + b2
\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=x^3\left(x-1\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2.\left(x^2+x+1\right)\)
x4 - x3 - x + 1
= (x4 - x3) - (x - 1)
= x3(x - 1) - (x - 1)
= (x3 - 1)(x - 1)
x^5+x^4+1
=x5+x4+x3+x2+x+1-x3-x2-x
=x3.(x2+x+1)+(x2+x+1)-x.(x2+x+1)
tự xử tiếp
\(x^{m+4}-x^{m+3}-x+1=x^{m+3}\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^{m+3}-1\right)\)
Ta có: \(x^{m+4}-x^{m+3}-x+1\)
\(=x^{m+3}\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^{m+3}-1\right)\)
\(x^4-2x^3+2x-1=x^3\left(x-1\right)-x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)=\left(x-1\right)\left(x^3-x^2-x+1\right)=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]=\left(x-1\right)^2\left(x^2-1\right)=\left(x-1\right)^3\left(x+1\right)\)
\(x^4-2x^3+2x-1\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)\)
\(=x^2\left(x^2+2x+1\right)+x+1\)
\(=x^2\left(x+1\right)^2+x+1\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+1\right]\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(x^4+2x^3+x^2+x+1\)
\(=x^2\left(x+1\right)^2+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+1\right)\)
\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)
\(A+x^4+64\)
\(A=\left(x^2\right)^2+8^2+16x^2-16x^2\)
\(A=\left(x^2+8\right)^2-16x^2\)
\(A=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(A=\left(x^2+8-4x\right)\left(x^2+8+4x\right)\)
\(x^4+64\\ \\=x^4+64+16x^2-16x^2\\ \\=\left(x^4+16x^2+64\right)-16x^2\\ \\=\left(x^2+8\right)^2-\left(4x\right)^2\\ \\=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)