Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

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

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

=> \(\left(x+y\right)+\left(x-y\right)=4+2\)

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)

 Bài giải:

Ta có: x(y + 3) + 2y(x + 2) = -15

<=> xy + 3x + 2xy + 2y = -15

<=>  3xy + 3x + 2y = -15

<=> 3x(y + 1) + 2(y + 1) – 2 = -15

<=> (y + 1)(3x + 2) = -15 + 2 = -13

Ta có bảng sau:

HỌC TỐT 

#MOON#

\(\left\{{}\begin{matrix}x+my=3\\x+2y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(m-2\right)y=2\\x=1-2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{2}{m-2}\\x=1-\dfrac{4}{m-2}=\dfrac{m-6}{m-2}\end{matrix}\right.\)

a, Ta có x < 0 ; y > 0 

\(x< 0\Rightarrow\dfrac{m-6}{m-2}< 0\)

Ta có : m - 2 > m - 6 

\(\left\{{}\begin{matrix}m-2>0\\m-6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>2\\m< 6\end{matrix}\right.\Leftrightarrow2< m< 6\)

\(y>0\Leftrightarrow\dfrac{2}{m-2}>0\Rightarrow m>2\)

Vậy 2 < m < 6 

b, \(x-2y=3\Rightarrow\dfrac{m-6}{m-2}-\dfrac{4}{m-2}=3\Leftrightarrow\dfrac{m-10}{m-2}=3\)

\(\Rightarrow m-10=3m-6\Leftrightarrow2m=-4\Leftrightarrow m=-2\)

Ta có:

\(\left|x+3\right|+\left|x-1\right|=\left|x+3\right|+\left|1-x\right|\ge\left|x+3+1-x\right|=4\)

\(3-y^2-2y=4-\left(y^2+2y+1\right)=4-\left(y+1\right)^2\le4\)

\(\Rightarrow\left|x+3\right|+\left|x-1\right|\ge3-y^2-2y\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x+3\right)\left(1-x\right)\ge0\\y+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le1\\y=-2\end{matrix}\right.\)

Các cặp số nguyên thỏa mãn là:

\(\left(x;y\right)=\left(-3;-2\right);\left(-2;-2\right);\left(-1;-2\right);\left(0;-2\right);\left(1;-2\right)\)

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

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

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

\(\Leftrightarrow y=\dfrac{x^3+x+1}{x+2}\)

-Vì \(x,y\) là các số nguyên nên:

\(\left(x^3+x+1\right)⋮\left(x+2\right)\)

\(\Rightarrow\left(x^3+2x^2-2x^2-4x+5x+10-9\right)⋮\left(x+2\right)\)

\(\Rightarrow\left[x^2\left(x+2\right)-2x\left(x+2\right)+5\left(x+2\right)-9\right]⋮\left(x+2\right)\)

\(\Rightarrow\left[\left(x+2\right)\left(x^2-2x+5\right)-9\right]⋮\left(x+2\right)\)

-Vì \(\left(x+2\right)\left(x^2-2x+5\right)⋮\left(x+2\right)\)

\(\Rightarrow9⋮\left(x+2\right)\)

\(\Rightarrow\left(x+2\right)\in\left\{1;3;9;-1;-3;-9\right\}\)

\(\Rightarrow x\in\left\{-1;1;7;-3;-5;-11\right\}\) (tmđk)

*Với \(x=-1\) thì \(y=\dfrac{\left(-1\right)^3+\left(-1\right)+1}{\left(-1\right)+2}=-1\) (tmđk)

*Với \(x=1\) thì \(y=\dfrac{1^3+1+1}{1+2}=1\)(tmđk)

*Với \(x=7\) thì \(y=\dfrac{7^3+7+1}{7+2}=39\)(tmđk)

*Với \(x=-3\) thì \(y=\dfrac{\left(-3\right)^3+\left(-3\right)+1}{\left(-3\right)+2}=29\)(tmđk)

*Với \(x=-5\) thì \(y=\dfrac{\left(-5\right)^3+\left(-5\right)+1}{\left(-5\right)+2}=43\)(tmđk)

*Với \(x=-11\) thì \(y=\dfrac{\left(-11\right)^3+\left(-11\right)+1}{\left(-11\right)+2}=149\)(tmđk)

Ta có: \(x^2y+3x^2-4y=15\)

=>\(x^2\left(y+3\right)-4y-12=15-12=3\)

=>\(\left(x^2-4\right)\left(y+3\right)=3\)

=>\(\left(x^2-4;y+3\right)\in\left\lbrace\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\rbrace\)

=>\(\left(x^2;y\right)\in\left\lbrace\left(5;0\right);\left(7;-2\right);\left(3;-6\right);\left(1;-4\right)\right\rbrace\)

mà x nguyên

nên \(\left(x^2;y\right)\in\left(1;-4\right)\)

=>(x;y)∈{(1;-4);(-1;-4)}