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Phân tích đa thức thành nhân tử:
\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
\(a^6-a^4+2a^3+2a^2\)
a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)
\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)
\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)
\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)
\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)
\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)
\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)
\(=4\left(2x-5\right)^2\left(x+1\right)\left(x+4\right)\)
b) \(a^6-a^4+2a^3+2a^2\)
\(=a^2\left(a^4-a^2+2a+2\right)\)
\(=a^2\left(a^4+a^3-a^3-a^2+2a+2\right)\)
\(=a^2\left[a^3\left(a+1\right)-a^2\left(a+1\right)+2\left(a+1\right)\right]\)
\(=a^2\left(a+1\right)\left(a^3-a^2+2\right)\)
Câu hỏi của giang ho dai ca - Toán lớp 8 - Học toán với OnlineMath
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y đúc như bài trên
\(a^4+8a^3+14a^2-8a-15\)
\(=a^4+8a^3+15a^2-a^2-8a-15\)
\(=a^2\left(a^2+8a+15\right)-\left(a^2+8a+15\right)\)
\(=\left(a^2+8a+15\right)\left(a^2-1\right)\)
\(=\left(a+3\right)\left(a+5\right)\left(a-1\right)\left(a+1\right)\)
a) 4(x2-y2)-8(x-ay)-4(a2-1)
=> 4x2-4y2-8x+8ay-4a2+4
=> 4(x2-y2-2x+2ay-a2+1)
c) a5+a4+a3 +a2 +a+1
=> a(a4+a3+a2+a+1)+1
\(x^8+x^4+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(x^5-x^4-1\)
\(=x^5-x^4+x^3-x^3+x^2-x-x^2+x-1\)
\(=\left(x^5-x^4+x^3\right)-\left(x^3-x^2+x\right)-\left(x^2-x+1\right)\)
\(=x^3\left(x^2-x+1\right)-x\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
=a^4(a+1)+a^2(a+1)+(a+1)
=(a+1)(a^4+a^2+1)
a5+a4+a3+a2+a+1
=a4(a+1)+a2(a+1)+(a+1)
=(a+1)(a4+a2+1)