4x2-12x-y2+2y+8
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\(a.\)
\(z^2-6z+5-t^2-4t\)
\(=z^2-6z+9-\left(t^2+4t+4\right)\)
\(=\left(z-3\right)^2-\left(t+2\right)^2\)
\(b.\)
\(4x^2-12x-y^2+2y+1\)
Câu này đề sai sao ấy em !
b, mik nghĩ đề sửa thành: \(4x^2-12x-y^2+2y+8\)
\(=4x^2-12x+9-y^2+2y-1\)
\(=\left(2x\right)^2-2.2.3.x+3^2-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)x^2-2x+2+4y^2+4y`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
`b)4x^2+y^2+12x+4y+13`
`=4x^2+12x+9+y^2+4y+4`
`=(2x+3)^2+(y+2)^2`
`c)x^2+17+4y^2+8x+4y`
`=x^2+8x+16+4y^2+4y+1`
`=(x+4)^2+(2y+1)^2`
`d)4x^2-12xy+y^2-4y+13`
`=4x^2-12x+9+y^2-4y+4`
`=(2x-3)^2+(y-2)^2`
a) \(x^2-2x+2+4y^2+4y=\left(x-1\right)^2+\left(2y+1\right)^2\)
b) \(4x^2+y^2+12x+4y+13=\left(2x+3\right)^2+\left(y+2\right)^2\)
c) \(x^2+17+4y^2+8x+4y=\left(x+4\right)^2+\left(2y+1\right)^2\)
d) \(4x^2-12x+y^2-4y+13=\left(2x-3\right)^2+\left(y-2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Sửa đề: \(x^2+3x+1\rightarrow x^2+2x+1\)
\(x^2+2x+1=\left(x+1\right)^2\)
b) \(x^2+y^2+2xy=\left(x+y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(-4x^2-9-12x=-\left(4x^2+12x+9\right)=-\left(2x+3\right)^2\)
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đây là hằng đẳng thức số 7 nhá bn
(x - 2y) (4x2 + 2xy + y2) = x3 - 8y3
\((x-2y)(4x^2+2xy+y^2 \\ =(x-2y)[(2x^2)+2x.y+y^2] \\ =x^3-(2y)^3 \\ =x^3-8y^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(d,15x^2y+20xy^2+25xy=5xy\left(3x+4y+5\right)\\ e,\left(x+y\right)^2-25=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\\ f,1-2y+y^2=\left(1-y\right)^2\\ h,4x^2+8xy-3x-6y=\left(4x^2+8xy\right)-\left(3x+6y\right)=4x\left(x+2y\right)-3\left(x+2y\right)=\left(x+2y\right)\left(4x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left(x+3\right)^2-4\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-4x+12\right)=0\\ \Leftrightarrow\left(x+3\right)\left(15-3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(b,=x^2\left(y-1\right)-\left(y-1\right)^2=\left(y-1\right)\left(x^2-y+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
\(4x^2-12x-y^2+2y+8\)
\(=\left(4x^2-12x+9\right)-\left(y^2-2y+1\right)\)
\(=\left(2x-3\right)^2-\left(y-1\right)^2\)
\(=\left(2x-3+y-1\right)\left(2x-3-y+1\right)\)
\(=\left(2x+y-4\right)\left(2x-y-2\right)\)