Question

[MâySadGirl]

đùng làm tắt nhé

Answer 1

[Kiều Vũ Linh]

a) \(2x^2-5x+2xy-5x-3=7\)

\(\left(2x^2+2xy\right)-\left(5x+5y\right)=7+3\)

\(2x\left(x+y\right)-5\left(x+y\right)=10\)

\(\left(x+y\right)\left(2x-5\right)=10\)

\(\Rightarrow2x-5\inƯ\left(10\right)=\left\{-10;-5;-2;-1;1;2;5;10\right\}\)

Mà \(x\in Z\)

\(\Rightarrow2x-5\) là số nguyên lẻ

\(\Rightarrow2x-5\in\left\{-5;-1;1;5\right\}\)

\(\Rightarrow2x\in\left\{0;4;6;10\right\}\)

\(\Rightarrow x\in\left\{0;2;3;5\right\}\)

*) \(x=0\)

\(\Rightarrow\left(0+y\right)\left(2.0-5\right)=10\)

\(\Rightarrow y.\left(-5\right)=10\)

\(\Rightarrow y=10:\left(-5\right)\)

\(\Rightarrow y=-2\)

*) \(x=2\Rightarrow\left(2+y\right)\left(2.2-5\right)=10\)

\(\Rightarrow\left(2+y\right).\left(-1\right)=10\)

\(\Rightarrow2+y=-10\)

\(\Rightarrow y=-10-2\)

\(\Rightarrow y=-12\)

*) \(x=3\Rightarrow\left(3+y\right)\left(2.3-5\right)=10\)

\(\Rightarrow3+y=10\)

\(\Rightarrow y=10-3\)

\(\Rightarrow y=7\)

*) \(x=5\Rightarrow\left(5+y\right)\left(2.5-5\right)=10\)

\(\Rightarrow\left(5+y\right).5=10\)

\(\Rightarrow5+y=10:5\)

\(\Rightarrow5+y=2\)

\(\Rightarrow y=2-5\)

\(\Rightarrow y=-3\)

Vậy ta tìm được các cặp giá trị \(\left(x;y\right)\) thỏa mãn:

\(\left(5;-3\right);\left(0;-2\right);\left(2;-12\right);\left(3;7\right)\)

Answer 2

[Kiều Vũ Linh]

b) \(2xy-3y^2+4x-6y-2=5\)

\(\left(2xy-3y^2\right)+\left(4x-6y\right)=5+2\)

\(y\left(2x-3y\right)+2\left(2x-3y\right)=7\)

\(\left(2x-3y\right)\left(y+2\right)=7\)

Bảng giá trị:

y+2	-7	-1	1	7
2x-3y	-1	-7	7	1
y	-9	-3	-1	5
x	-14	-8	2	8

Vậy ta tìm được các cặp giá trị \(\left(x;y\right)\) thỏa mãn:

\(\left(-14;-9\right);\left(-8;-3\right);\left(2;-1\right);\left(8;5\right)\)