(x+3y)^2-(2x-3y)^2-2x^2+12y^2
Những câu hỏi liên quan
tìm x;y
a) 4x2+13y+12xy−18y−4x+104x2+13y+12xy−18y−4x+10
b) 4x2+12xy+9y2+4y2−18y−4x+104x2+12xy+9y2+4y2−18y−4x+10
c) (2x+3y)2−2(2x+3y)+1+4y2−12y+9(2x+3y)2−2(2x+3y)+1+4y2−12y+9
d) (2x+3y−1)+(2y−3)2=0
b.x^2 +6x - 3 (x+6)
c. 2x^3y-8x^2y+8xy
d.y^2-x^2-12y+36
\(b,x^2+6x-3\left(x+6\right)=x\left(x+6\right)-3\left(x+6\right)=\left(x+6\right)\left(x-3\right)\\ c,2x^3y-8x^2y+8xy=2xy\left(x^2-4x+4\right)=2xy\left(x-2\right)^2\\ d,y^2-x^2-12y+36=\left(y^2-12y+36\right)-x^2=\left(y-6\right)^2-x^2=\left(y-x-6\right)\left(y+x-6\right)\)
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Phân tích đa thức thành nhân tử
1/ x2 - 3xy - 2y2
2/ 2x2 + 5xy + y2
3/ 6a2 - ab - 2b2 + a + 4b - 2
4/ 2x2 + 5x - 12y2 + 12y - 3 - 10xy
5/ 2x2 - 7xy + x + 3y2 - 3y
6/ 4x2 - 4xy - 3y2 - 2x + 3y
7/ 3x2 - 5xy + 2y2 + 4x - 4y
phân tích đa thức sau thành nhân tử
a\(12x^3y-24x^2y^2+12xy^3\)
b\(x^2-6x+xy-6y\)
c\(2x^2+2xy-x-y\)
d\(ax-2x-a^2+2a\)
e\(x^3-3x^2+3x-1\)
f\(3x^2-3y^2-12x-12y\)
b: \(x^2-6x+xy-6y\)
\(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
c: \(2x^2+2xy-x-y\)
\(=2x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(2x-1\right)\)
e: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
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Tìm x,y,z biết:
a) x2+4y2+z2=2x+12y-4z-14
b) x2+3y2+2z2-2x+12y+4z+15=0
Tìm x,y,z bik
a)\(x^2+4y^2+z^2=2x+12y-4z-14\)
b) \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
a. \(x^2+4y^2+z^2=2x+12y-4z-14\)
\(\Leftrightarrow x^2+4y^2+z^2-2x-12y+4z+14=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\\\left(2y-3\right)^2\ge0\\\left(z+2\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\2y-3=0\\z+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
b. \(x^2+3y^2+2z^2-2x+12y+4z+15=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)+3\left(y^2+4y+4\right)+2\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=-1\end{matrix}\right.\)
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biết \(x^2+3y^2-2x+12y+13\) . khi đó x=... y= .....
Ta có:x2+3y2-2x+12y+13=0<=>x2+3y2-2x+12y+1+12=0
<=>(x2-2x+1)+(3y2+12y+12)=0<=>(x-1)2+3(y+2)2=0
Vì (x-1)2\(\ge0\);3(y+1)2\(\ge0\) nên:(x-1)2+3(y+2)2\(\ge0\)
Dấu "=" xảy ra khi:\(\begin{cases} (x-1)^2=0\\ 3(y+2)^2=0 \end{cases}\)<=>\(\begin{cases} x-1=0\\ y+2=0 \end{cases}\)<=>\(\begin{cases} x=1\\ y=-2 \end{cases}\)
Vậy x=1;y=-2
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xin loi mk ghi thieu
x2+3y2-2x+12y+13=0
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biết \(x^2-3y^2-2x+12y+13=0\) khi đó x= ......... y= ..........
\(x^2-3y^2-2x+12y+13=0\)
\(\Rightarrow\left(x^2-2x+1\right)-3\left(y^2-4y+4\right)+4^2=0\)HÌnh như hơi vô lý bạn ạg
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Giải HPT
\(\left\{{}\begin{matrix}2x^2-xy+3y^2=7x+12y-1\\x-y+1=0\end{matrix}\right.\)
\(PT\left(2\right)\Leftrightarrow x=y-1\\ PT\left(1\right)\Leftrightarrow2\left(y-1\right)^2+y\left(1-y\right)+3y^2=7\left(y-1\right)+12y-1\\ \Leftrightarrow2y^2-11y+5=0\\ \Leftrightarrow\left[{}\begin{matrix}y=5\Leftrightarrow x=4\\y=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
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