Phân tích đa thức sau thành nhân tử: \(x^3\left(x^2-7\right)^2-36x\)
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Những câu hỏi liên quan
LN
3
DN
1
4 tháng 9 2021
\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)
\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)
\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)
\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)
\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)
\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)
KN
8 tháng 7 2019
a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)
\(=\left(x^4-2x^3+5x^2-4x+4\right)+\left(x^2-4x+4\right)\)
\(=x^4-2x^3+6x^2-8x+8\)
\(=\left(x^4-2x^3+2x^2\right)+\left(4x^2-8x+8\right)\)
\(=x^2\left(x^2-2x+2\right)+4\left(x^2-2x+2\right)\)
\(=\left(x^2+4\right)\left(x^2-2x+2\right)\)
B
8 tháng 7 2019
\(x^4-9x^3+28x^2-36x+16\)
\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)
\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)
\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)
\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)
\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)
\(=\left(x-1\right)\left(x-2\right)\left(x^2-4x-2x+8\right)\)
\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)
\(=\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\left(x-4\right)\)
DN
4
1 tháng 9 2021
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
26 tháng 10 2018
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
\(x^3\left(x^2-7\right)^2-36x\)
\(=x.\left[x^2.\left(x^2-7\right)^2-36\right]\)
\(=x.\left[\left(x^3-7x\right)^2-6^2\right]\)
\(=x.\left(x^3-7x-6\right).\left(x^3-7x+6\right)\)
\(=x.\left(x+1\right)\left(x^2-x-6\right).\left(x-1\right).\left(x^2+x-6\right)\)
\(=x.\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x-1\right).\left(x-2\right).\left(x-3\right)\)
Ta có : \(x^3\left(x^2-7\right)^2-36x\)
= \(x^3\left(x^4-14x^2+49\right)-36x\)
= \(x\left(x^6-14x^4+49x^2-36\right)\)
= \(x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)\)---- chỗ này tắt
= (x-3)(x-2)(x-1)x(x+1)(x+2)(x+3)