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a. Ta có: x2+y2-2x+4y+5=0
⇌(x-1)2+(y-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
b. Ta có: 4x2+y2-4x-6y+10=0
⇌ (2x-1)2+(y-3)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\end{matrix}\right.\)
c.Ta có: 5x2-4xy+y2-4x+4=0
⇌(2x-y)2+(x-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=2\end{matrix}\right.\)
d.Ta có: 2x2-4xy+4y2-10x+25=0
⇌ (x-2y)2+(x-5)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{2}\\x=5\end{matrix}\right.\)
\(A,=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\\ B,=25-\left(4x^2+4xy+y^2\right)=25-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)
Mình làm và sửa đề đúng luôn nhé !
1) \(36x^2-a^2+10a-25\)
\(=\left(6x\right)^2-\left(a^2-10a+25\right)\)
\(=\left(6x\right)^2-\left(a-5\right)^2\)
\(=\left(6x-a+5\right)\left(6x+a-5\right)\)
2) \(4x^2-4xy+y^2-25a^2+10a-1\)
\(=\left(2x-y\right)^2-\left(5a-1\right)^2\)
\(=\left(2x-y-5a+1\right)\left(2x-y+5a-1\right)\)
3) \(m^2-6m+9-x^2+4xy-4y^2\)
\(=\left(m-3\right)^2-\left(x-2y\right)^2\)
\(=\left(m-3-x+2y\right)\left(m+3-x+2y\right)\)
\(a,=2xy\left(2y-x\right)\\ b,=x^2\left(x-4\right)+5\left(x-4\right)=\left(x^2+5\right)\left(x-4\right)\\ c,=\left(x-y\right)\left(x^2-25\right)=\left(x-y\right)\left(x-5\right)\left(x+5\right)\)
phân tích các đa thức sau thành nhân tử:
25-4x^2-4xy-y^2
x^2+2xy+y^2-xz-yz
x^2-4xy+4y^2-z^2+4zt-4t^2
Giải:
a) \(25-4x^2-4xy-y^2\)
\(=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2\)
\(=\left(5-2x-y\right)\left(5+2x+y\right)\)
Vậy ...
b) \(x^2+2xy+y^2-xz-yz\)
\(=x^2+2xy+y^2-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
Vậy ...
c) \(x^2-4xy+4y^2-z^2+4zt-4t^2\)
\(=\left(x^2-4xy+4y^2\right)-\left(z^2-4zt+4t^2\right)\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)
\(=\left(x-2y+z-2t\right)\left(x-2y-z+2t\right)\)
Vậy ...
phân tích các đa thức sau thành nhân tử:
1, \(25-4x^2-4xy-y^2\)
\(=5^2-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2\)
\(=\left(5-2x-y\right)\left(5+2x+y\right)\)
2,\(x^2+2xy+y^2-xz-yz\)
\(=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
3,\(x^2-4xy+4y^2-z^2+4zt-4t^2\)
\(=\left(x^2-4xy+4y^2\right)-\left(z^2-4zt+4t^2\right)^{ }\)
\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)
\(=\left(x-2y-z+2t\right)\left(x-2y+z-2t\right)\)
\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)
\(a,x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
\(b,25-4x^2-4xy-y^2\)
\(=25-\left(4x^2+4xy+y^2\right)\)
\(=5^2-\left(2x+y\right)^2\)
\(=\left(5-2x+y\right)\left(5+2x+y\right)\)
\(c,x^3-x+y^3-y\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+1\right)\)
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