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\(M=8x^2-2xy-y^2-5x^2+2xy+3y^2=3x^2+2y^2>=0\forall x,y\)
Bài 1 Tớ giải từng bài nhé ! Ko có ý đồ câu điểm.
\(A=4x^2-5xy+xy^2\)
\(B=3x^2+2xy-xy^2\)
Ta có : \(A+B=4x^2-5xy+xy^2+3x^2+2xy-xy^2\)
\(=7x^2-3xy\)
\(A-B=4x^2-5xy+xy^2-3x^2-2xy+xy^2\)
\(=x^2-7xy+2xy^2\)
Bài 2 : N ở đâu ?
Ta có : \(M+\left(5x^2-2xy\right)=xy^2+xy^3-y^2\)
\(M=xy^2+xy^3-y^2-5x^2+2xy\)
Bài 3 :
\(A=x^2y-xy^2+xy^2=x^2y\)
\(B=xy+4xy^2-2x-1\)
a) (-xy)10 : (-xy)5 = (-xy)2
b) (5x2y4) : 10x2y = y3/2
c) (15x4y3z2) : (5x2y2z2) = 3x2y
d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}x^2y^2\right)=\frac{\frac{3}{4}x^3y^2}{-\frac{1}{2}x^2y^2}=-\frac{3}{2}x\)
e) 6x3y5 : 12x3y2 = y3/2
f) (25x5 - 5x4 + 10x2) : (5x2)
= 5x3 - x2 + 2
g) \(\frac{2}{3}xy\cdot\left(2x^2y-3xy+y^2\right)=\frac{4}{3}x^3y^2-2x^2y^2+\frac{2}{3}xy^3\)
h) \(\frac{3}{4}x^3y^5z:\left(5x^2y^2z\right)=\frac{\frac{3}{4}x^3y^5z}{5x^2y^2z}=\frac{3}{20}xy^3\)
\(C=x^3+x^2y-xy^3-y^4+x^2-y^3+3=\left(x^3+x^2y+x^2\right)-\left(xy^3+y^4+y^3\right)+3=x^2\left(x+y+1\right)-y^3\left(x+y+1\right)+3=x^2.0+y^3.0+3=0+0+3=3\)
\(Taco:\left\{{}\begin{matrix}\left(x-2\right)^4\ge0\forall x\\\left(2y-1\right)^{2014}\ge0\forall y\end{matrix}\right.mà:\left(x-2\right)^4+\left(2y-1\right)^{2014}\le0\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^4=0\\\left(2y-1\right)^{2014}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\frac{1}{2}\end{matrix}\right.\Rightarrow D=21x^2y+4xy^2=xy\left(21x+4y\right)=\frac{2}{2}\left(42+2\right)=44\)
\(Bài4\)
\(xy+3x-y=6\Leftrightarrow xy+3x-y-3=3\Leftrightarrow x\left(y+3\right)-\left(y+3\right)=3\Leftrightarrow\left(x-1\right)\left(y+3\right)=3;x\in Z\Rightarrow x-1\in Z\Rightarrow x-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)
\(+,x-1=-1\Rightarrow\left\{{}\begin{matrix}x=0\\y+3=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\left(thoaman\right)\)
\(+,x-1=-3\Rightarrow\left\{{}\begin{matrix}x=-2\\y+3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-4\end{matrix}\right.\left(thoaman\right)\)
\(+,x-1=3\Rightarrow\left\{{}\begin{matrix}x=4\\y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-2\end{matrix}\right.\left(thoaman\right)\)
\(+,x-1=1\Rightarrow\left\{{}\begin{matrix}x=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\left(thoaman\right)\)
\(Vậy:\left(x,y\right)\in\left\{\left(2;0\right);\left(4;-2\right);\left(-2;-4\right);\left(0;-6\right)\right\}\)