Phân tích đa thức sau thành nhân tử
x2-9-4(x+3)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^2-4-3\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
\(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x-4\right)\left(x+1\right)\)
\(x^2-11x+3\\ =\left(x^2-4x+4\right)-7x-1\\ =\left(x-2\right)^2-\left(\sqrt{7x+1}\right)^2\\ =\left(x-2-\sqrt{7x+1}\right)\left(x-2+\sqrt{7x+1}\right)\)
\(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left[\left(x-y\right)^2-1^2\right]+\left(3x-3y-3\right)\)
\(=\left[\left(x-y\right)-1\right]\left[\left(x-y\right)+1\right]+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left[\left(x-y+1\right)+3\right]\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
\(=x^2+x-6x+6\\ =x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x+6\right)\)
\(x^2-6x+7=x^2-6x+9-2\\ =\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\\ x^4+64=x^4+16x^2+64-16x^2\\ =\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\\ a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2\\ =\left(a^2+2b^2\right)^2-4a^2b^2\\ =\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)
= x2 -7x -x +7
= x. (x-7) - (x-7)
= (x-1)(x-7)
Chúc bạn học tốt nha!
`x^2 -4x+4-y^2`
`=(x^2 -4x+4)-y^2`
`=(x-2)^2 -y^2`
`=(x-2-y)(x-2+y)`
`x^2+2xy+y^2-x-y`
`=(x^2+2xy+y^2) -(x+y)`
`=(x+y)^2 -(x+y)`
`=(x+y)(x+y-1)`
`x^2-2xy+y^2-9`
`=(x^2-2xy+y^2)-3^2`
`=(x-y)^2-3^3`
`=(x-y-3)(x-y+3)`
Tách ra đi cậu.
\(x^2-9-4\left(x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3-4\right)\)
\(=\left(x+3\right)\left(x-7\right)\)
\(x^2-9-4\left(x+3\right)\)
\(=\left(x^2-9\right)-4\left(x+3\right)\)
\(=\left(x^2-3^2\right)-4\left(x+3\right)\)
\(=\left(x-3\right)\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3-4\right)\)
\(=\left(x+3\right)\left(x-7\right)\)