Phân tích đa thức thành nhân tử:
\(\text{x^4 + 5x^3 + 10x - 4 }\)(Bằng phương pháp thêm bớt hạng tử)
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x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
x5-x4-1=x5-x3-x2-x4+x2+x+x3-x-1
=x2.(x3-x-1)-x.(x3-x-1)+(x3-x-1)
=(x3-x-1)(x2-x+1)
x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)
k mk nha
\(x^4+1\)
\(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-2x^2\)
\(=\left(x^2-\sqrt{2}x+1\right)\left(x^2+\sqrt{2}x+1\right)\)
\(x^8+x^4+1\)
\(=x^4.\left(x^4+1\right)+\left(x^4+1\right)-x^4\)
\(=\left(x^4+1\right).\left(x^4+1\right)-\left(x^2\right)^2\)
\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right).\left(x^4+1+x^2\right)\)
\(=x^4+2x^2+1-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2+1-\sqrt{2}x\right)\left(x^2+1+\sqrt{2}x\right)\)
\(x^4+1\)
\(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-\left(x\sqrt{2}\right)^2\)
\(=\left(x^2-x\sqrt{2}+1\right)\left(x^2+x\sqrt{2}+1\right)\)
\(x^4+81\)
\(=x^4+3^4\)
\(=\left(x^2+3^2\right)^2-2x^23^2\)
\(=\left(x^2+\sqrt{2}x3+3^2\right)\left(x^2-\sqrt{2}x3+3^2\right)\)
nguồn gg
\(x^4+81\)
\(=x^4+18x^2+81-18x^2\)
\(=\left(x^2+9\right)^2-18x^2\)
\(=\left(x^2-3\sqrt{2}x+9\right)\left(x^2+3\sqrt{2}x+9\right)\)
\(x^3+x^2+4\)
\(=\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(2x+4\right)\)
\(=x^2.\left(x+2\right)-x.\left(x+2\right)+2.\left(x+2\right)\)
\(=\left(x+2\right).\left(x^2-x+2\right)\)
k đi rồi giúp cho
k rồi giúp cho