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a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
1) \(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
Phân tích đa thức sau thành nhân tử :
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=a^3\left(b-c\right)+b^3c-b^3a+c^3a-c^3b\\ \)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b^2-c^2\right)-a\left(b^3-c^3\right)\)
\(\Rightarrow\)\(a^3\left(b-c\right)+bc\left(b-c\right)\left(b+c\right)-a\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+bc\left(b+c\right)-a\left(b^2+bc+c^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(a^3+b^2c+bc^2-ab^2-abc-ac^2\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(bc\left(c-a\right)+b^2\left(c-a\right)-a\left(c^2-a^2\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-a\left(c+a\right)\right)\)
\(\Rightarrow\)\(\left(b-c\right)\left(c-a\right)\left(bc+b^2-ac-a^2\right)\)
\(\left(b-c\right)\left(c-a\right)\left(b^2-a^2+c\left(b-a\right)\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\left(a+b+c\right)\)
\(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2-a^3-b^3-c^3+4abc\)
\(=a\left(b-c\right)^2-a^3+4abc+b\left(c-a\right)^2-b^3+c\left(a-b\right)^2-c^3\)
\(=a\left[\left(b-c\right)^2+4bc-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left[\left(b+c\right)^2-a^2\right]+b\left[\left(c-a\right)^2-b^2\right]+c\left[\left(a-b\right)^2-c^2\right]\)
\(=a\left(b+c+a\right)\left(b+c-a\right)+b\left(c-a+b\right)\left(c-a-b\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[a\left(b+c+a\right)+b\left(c-a-b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[ab+ac+a^2+bc-ab-b^2\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left[c\left(a+b\right)+\left(a-b\right)\left(a+b\right)\right]+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(b+c-a\right)\left(a+b\right)\left(a-b+c\right)+c\left(a-b+c\right)\left(a-b-c\right)\)
\(=\left(a-b+c\right)\left[b^2-\left(a-c\right)^2\right]\)
\(=\left(a-b+c\right)\left(b+a-c\right)\left(b-a+c\right)\)
a)hiệu hai lập phương
b) tổng hai lập phương
hằng đẳng thức có sẵn
bạn giải ra hộ mk đi