Phân tích đa thức thành nhân tử
a) a^6 - a^4 + 2a^3 + 2a^2
b) ( a + b )^3 - ( a - b )^3
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Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
`a)x^4+2x^2y+y^2`
`=(x^2+y)^2`
`b)(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a)(2a+b+2b+a)`
`=(a-b)(3a+3b)`
`=3(a-b)(a+b)`
`c)8a^3-27b^3-2a(4a^2-9b^2)`
`=(2a-3b)(4a^2+6ab+9b^2)-2a(2a-3b)(2a+3b)`
`=(2a-3b)(4a^2+6ab+9b^2-3a^2-6ab)`
`=9b^2(2a-3b)`
a) Ta có: \(x^4+2x^2y+y^2\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot y+y^2\)
\(=\left(x^2+y\right)^2\)
b) Ta có: \(\left(2a+b\right)^2-\left(2b+a\right)^2\)
\(=\left(2a+b-2b-a\right)\left(2a+b+2b+a\right)\)
\(=\left(a-b\right)\left(3a+3b\right)\)
\(=3\left(a+b\right)\left(a-b\right)\)
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)
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a(a+2b)3 -b(2a+b)3
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left[\left(a^2\right)^2+ \left(b^2\right)^2\right]-2ab\left(a^2-b^2\right)\)
\(=\left(a^2+b^2\right)\left(a^2-b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a^2-b^2\right)\left(a^2-2ab+b^2\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a-b\right)^2\)
\(=\left(a-b\right)^3\left(a+b\right)\)
\(a.\left(a+2b\right)^3-b.\left(2a+b\right)^3\)
\(=a.\left(a+20+b\right)^3-b.\left(20+a+b\right)^3\)
\(=\left(a-b\right).\left(a+20+b\right)^3\)
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 ?