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cái này mk chưa hok tới!!!
54746746745764565465476467568457879797689685856
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a, \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2=\left[\left(x+1\right)-\left(y-3\right)\right]^2\)
\(=\left(x+1-y+3\right)^2=\left(x-y+4\right)^2\)
b, \(a^2+b^2+2a-2b-2ab=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left[\left(a-b\right)+2\right]=\left(a-b\right)\left(a-b+2\right)\)
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Gọi a+b =x có:
a(x+b)3−b(x+a)3
=a(x3+3x2b+3xb2+b3)−b(x3+3x2a+3xa2+a3)
=ax3+3ax2b+3axb2+ab3−bx3−3bx2a−3bxa2−ba3
=(a−b)x3+(3ax2b−3bx2a)+(3axb2−3bxa2)+ab3−ba3
=(a−b)x3+3axb(b−a)+ab(b2−a2)
=−x3(b−a)+3axb(b−a)+ab(b+a)(b−a)
=−x3(b−a)+3axb(b−a)+(a2b+ab2)(b−a)
=(b−a)(−x3+3axb+a2b+ab2)
nho lik e
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\(2\left(x-3\right)^3-4\left(3-x\right)^2-x+3\)
\(=2\left(x-3\right)^3-4\left(x-3\right)^2-\left(x-3\right)\)
\(=\left(x-3\right)\left[2\left(x-3\right)^2-4\left(x-3\right)-1\right]\)
\(=\left(x-3\right)\left[2\left(x^2-6x+9\right)-4x+12-1\right]\)
\(=\left(x-3\right)\left[2x^2-12x+18-4x+11\right]\)
\(=\left(x-3\right)\left[2x^2-16x+29\right]\)
\(ab\left(a-b\right)-2a+2b\)
\(=ab\left(a-b\right)-2\left(a-b\right)\)
\(=\left(ab-2\right)\left(a-b\right)\)
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a(a+2b)3 -b(2a+b)3
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left[\left(a^2\right)^2+ \left(b^2\right)^2\right]-2ab\left(a^2-b^2\right)\)
\(=\left(a^2+b^2\right)\left(a^2-b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-b\right)^2\)
\(=\left(a-b\right)^3\left(a+b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(a.\left(a+2b\right)^3-b.\left(2a+b\right)^3\)
\(=a.\left(a+20+b\right)^3-b.\left(20+a+b\right)^3\)
\(=\left(a-b\right).\left(a+20+b\right)^3\)
Thế này có phải là phân tích đa thức thành nhân tử k ạ
Chúc bạn học tốt
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=\left(a^4+6a^3b+12a^2b^2+8ab^3\right)-\left(b^4+8a^3b+12a^2b^2+6ab^3\right)\)
\(=a^4-b^4-2a^3b+2ab^3\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)^3\left(a+b\right)\)
OK ?