\(\left(2x+5\right)\left(y-3\right)=22\)

\(\Rightarrow\left(2x+5\right);\left(y-3\right)\in\left\{1;2;11;22\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-2;25\right);\left(-\dfrac{3}{2};14\right);\left(3;5\right);\left(\dfrac{17}{2};4\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(3;5\right)\right\}\left(\left(x;y\inℤ^+\right)\right)\)

\(\left(4x-1\right)\left(y-3\right)=18\)

\(\Rightarrow\left(4x-1\right);\left(y-3\right)\in U\left(18\right)=\left\{1;2;3;6;9;18\right\}\left(x,y\inℤ^+\right)\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{1}{2};31\right);\left(\dfrac{3}{4};12\right);\left(1;9\right);\left(\dfrac{7}{4};6\right);\left(\dfrac{5}{2};5\right);\left(\dfrac{19}{4};4\right)\right\}\left(x,y\inℤ^+\right)\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;9\right)\right\}\left(x,y\inℤ^+\right)\)

\(\left(2x-1\right)\left(y-7\right)=22\)

\(\Rightarrow\left(2x-1\right);\left(y-7\right)\in\left\{1;2;11;22\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;29\right);\left(\dfrac{3}{2};18\right);\left(6;9\right);\left(\dfrac{23}{2};8\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;29\right);\left(6;9\right)\right\}\left(x;y\inℤ^+\right)\)

cách kiếm xu kiểu gì

a: \(x^2+y^2(x-y+1)-\left(x-1\right)\cdot y=22\)

=>\(x^2 + y^2(x - y + 1) - xy + y = 22\)

=>\(x^2-xy+y^2(x-y+1)+y=22\)

=>\(x(x-y+1)-x+y^2(x-y+1)+y=22\)

\(\iff(x-y+1)(x+y^2)-(x-y)=22\)

\(\iff(x-y+1)(x+y^2)-(x-y+1)+1=22\)

=>\((x-y+1)(x+y^2-1)=21\)

TH1: \(\begin{cases}x-y+1=1\\ x+y^2-1=21\end{cases}\)

=>\(\begin{cases}x=y\\ y+y^2-1=21\end{cases}\iff y^2+y-22=0\)

=>\(y^2+y+\frac14-\frac{89}{4}=0\)

=>\(\left(y+\frac12\right)^2=\frac{89}{4}\)

=>\(\left[\begin{array}{l}y+\frac12=\frac{\sqrt{89}}{2}\\ y+\frac12=-\frac{\sqrt{89}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}y=\frac{\sqrt{89}-1}{2}\left(loại\right)\\ y=\frac{-\sqrt{89}-1}{2}\left(loại\right)\end{array}\right.\)

=>Loại

TH2: \(\begin{cases}x-y+1=21\\ x+y^2-1=1\end{cases}\iff\begin{cases}x=y+20\\ (y+20)+y^2-1=1\end{cases}\iff y^2+y+18=0\)

=>\(y^2+y+\frac14+\frac{71}{4}=0\)

=>\(\left(y+\frac12\right)^2+\frac{71}{4}=0\) (vô lý)

=>Loại

TH3: \(\begin{cases}x-y+1=3\\ x+y^2-1=7\end{cases}\iff\begin{cases}x=y+2\\ (y+2)+y^2-1=7\end{cases}\iff y^2+y-6=0\)

=>(y+3)(y-2)=0

=>y=-3 hoặc y=2

Nếu y=-3 thì x-y+1=3

=>x-y=2

=>x-(-3)=2

=>x+3=2

=>x=-1

Nếu y=2 thì x-y+1=3

=>x-2+1=3

=>x-1=3

=>x=4

TH4: \(\begin{cases}x-y+1=7\\ x+y^2-1=3\end{cases}\iff\begin{cases}x=y+6\\ (y+6)+y^2-1=3\end{cases}\iff y^2+y+2=0\)

=>y∈∅

=>Loại

bai toan nay kho

Câu 1

1abc= 1000+abc

Ta có abc chia hết cho abc

Suy ra số dư của 1000:abc cũng là số dư của 1abc:abc

Suy ra 1000:abc dư 3

Suy ra 997:abc

Mà abc là số có 3 chữ số nên abc=997

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

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

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

a: =>-2x=90/91

hay x=-45/91

b: =>2x=-7

hay x=-7/2

c: ->-3x=-12

hay x=4