Question

[channel Anhthư]

Bài 1, Chứng minh rằng:

Q(x)=-Q(-x)

Bài 2 : Tính giá trị biểu thứ :

C= x³+x²y-xy³-y⁴+x²-y³+3 biết x+y+1=0

D=21x²y+4xy²với (x-2)⁴+(2y-1) ²⁰¹⁴ <(hoặc bằng)0

Bài 4 : Tìm cặp số nguyên x, y thỏa mãn :

xy+3x-y=6

Answer 1

[svtkvtm]

\(C=x^3+x^2y-xy^3-y^4+x^2-y^3+3=\left(x^3+x^2y+x^2\right)-\left(xy^3+y^4+y^3\right)+3=x^2\left(x+y+1\right)-y^3\left(x+y+1\right)+3=x^2.0+y^3.0+3=0+0+3=3\)

\(Taco:\left\{{}\begin{matrix}\left(x-2\right)^4\ge0\forall x\\\left(2y-1\right)^{2014}\ge0\forall y\end{matrix}\right.mà:\left(x-2\right)^4+\left(2y-1\right)^{2014}\le0\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^4=0\\\left(2y-1\right)^{2014}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\2y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\frac{1}{2}\end{matrix}\right.\Rightarrow D=21x^2y+4xy^2=xy\left(21x+4y\right)=\frac{2}{2}\left(42+2\right)=44\)

\(Bài4\)

\(xy+3x-y=6\Leftrightarrow xy+3x-y-3=3\Leftrightarrow x\left(y+3\right)-\left(y+3\right)=3\Leftrightarrow\left(x-1\right)\left(y+3\right)=3;x\in Z\Rightarrow x-1\in Z\Rightarrow x-1\inƯ\left(3\right)=\left\{-1;1;-3;3\right\}\)

\(+,x-1=-1\Rightarrow\left\{{}\begin{matrix}x=0\\y+3=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-6\end{matrix}\right.\left(thoaman\right)\)

\(+,x-1=-3\Rightarrow\left\{{}\begin{matrix}x=-2\\y+3=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-4\end{matrix}\right.\left(thoaman\right)\)

\(+,x-1=3\Rightarrow\left\{{}\begin{matrix}x=4\\y+3=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-2\end{matrix}\right.\left(thoaman\right)\)

\(+,x-1=1\Rightarrow\left\{{}\begin{matrix}x=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\left(thoaman\right)\)

\(Vậy:\left(x,y\right)\in\left\{\left(2;0\right);\left(4;-2\right);\left(-2;-4\right);\left(0;-6\right)\right\}\)