Question

Tìm 2 số nguyên x,y thỏa mãn :

a/ \(\left(x+y+1\right)^2=3\left(x^2+y^2\right)+1\)

b/ \(\left(2x-y-2\right)^2=7\left(x-2y-y^2-1\right)\)

c/ \(x^2+6y-4xy+4x-15y=0\)

Answer 1

Xin câu a :3

a) (x + y + 1)² = 3(x² + y²) + 1

<=> x² + y² + 1 + 2xy + 2x + 2y = 3x² + 3y² + 1

<=> 2x² + 2y² - 2xy - 2x - 2y = 0

<=> (x² - 2xy + y²) + (x² - 2x + 1) + (y² - 2y + 1) = 2

<=> (x - y)² + (x - 1)² + (y - 1)² = 2

Vì 2 = 0² + 1² + 1² nên ta có các TH sau:

TH1:

\(\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-1\right)^2=1\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=y=2\\x=y=0\end{matrix}\right.\)

TH2:

\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=0\\\left(y-1\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1;y=0\\x=1;y=2\end{matrix}\right.\)

TH3:

\(\left\{{}\begin{matrix}\left(x-y\right)^2=1\\\left(x-1\right)^2=1\\\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=0;y=1\end{matrix}\right.\)

Vậy ...

Answer 2

a) ta có : \(\left(x+y+1\right)^2=3\left(x^2+y^2\right)+1\)

\(\Leftrightarrow x^2+y^2+1+2xy+2y+2x=3x^2+3y^2+1\)

\(\Leftrightarrow-\left(x-1\right)^2-\left(y-1\right)^2=\left(x-y\right)^2-2\le0\)

\(\Leftrightarrow-\sqrt{2}\le x-y\le\sqrt{2}\) --> ...

b) \(\left(2x-y-2\right)^2=7\left(x-2y-y^2-1\right)\)

\(\Leftrightarrow4x^2+y^2+4-4xy+4y-4x=7x-14y-7y^2-7\)

\(\Leftrightarrow2x^2-4xy+2y^2+2x^2-11x+\dfrac{121}{16}+6y^2+18y+\dfrac{9}{4}=\dfrac{-19}{16}\left(vl\right)\)

câu c tương tự .

Answer 3

xin câu b,

giải pt

\(4x^2-x\left(15+4y\right)+8y^2+18y+11=0\)

\(\Delta=\left(15+4y\right)^2-4.4.\left(8y^2+18y+11\right)=-112y^2-168y+49\ge0\)

\(\Rightarrow\left(y+\dfrac{3}{4}\right)^2\le1\)

\(\Rightarrow-\dfrac{7}{4}\le y\le\dfrac{1}{4}\)

vậy y=0 hoặc bằng -1

thay vào tìm x

Answer 4

Mysterious Person DƯƠNG PHAN KHÁNH DƯƠNG Phùng Khánh Linh Khôi Bùi help :>

Answer 5

Slot,cho t slot,thằng nào động vào t chém =))

Answer 6

Cứ Δ mà quất, khó quá bỏ qua :))