Ta có bảng 

(x+3).(y-5)=25

Ta có:các số nhân nhau bằng 25 là:5x5

=>x+3=5  và  y-5=5

    x=5-3        y=5+5

    x=2          y=10

Vậy:x=2;y=10    chúc bạn học tốt

số nguyên mà 

25 – 2x = 16 – 3x

-2x + 3x = 16 - 25

x = -9

Vậy x = -9

x - (47 - 22 ) = 5 + (10-4x)

x - 47 + 22    = 5 + 10 - 4x

x + 4x = 5 + 10 - 22 + 47

5x       = 40

x         = 40 : 5

x         = 8

Vậy x = 8.

~ HOK TỐT ~

\(\text{25 – 2x = 16 – 3x}\)

\(2x+3x=16-25\)

\(5x=-9\)

\(\Rightarrow x=\frac{-9}{5}\)

\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)

\(x-47+22=5+10-4x\)

\(x+4x=5+10-22+47\)

\(5x=40\)

\(\Rightarrow x=8\)

học tốt

\(25-2x=16-3x\)

\(\Leftrightarrow25-2x-16=-3x\)

\(\Leftrightarrow-2x+9=-3x\)

\(\Leftrightarrow-3x+2x=9\)

\(\Leftrightarrow-x=9\Leftrightarrow x=9\)

Vậy ...

\(x-\left(47-22\right)=5+\left(10-4x\right)\)

\(\Leftrightarrow x-47+22=5+10-4x\)

\(\Leftrightarrow x-25=5+10-4x\)

\(\Leftrightarrow x-25=-4x+15\)

\(\Leftrightarrow x=-4x+15+25\)

\(\Leftrightarrow x=-4x+40\)

\(\Leftrightarrow x+4x=40\Leftrightarrow5x=40\Leftrightarrow x=8\)

Vậy ...

(x - 2)(y + 3) = 7

\(\Rightarrow\) x - 2 và y + 3 \(\in\) Ư₍₇₎

Ư₍₇₎ = {1; -1; 7; -7}

Xét các TH:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=7\\y+3=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=9\\y=-4\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-7\\y+3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-5\\y=-2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=1\\y+3=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=-1\\y+3=-7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-10\end{matrix}\right.\end{matrix}\right.\)

Vậy x \(\in\) {9; -5; 3; 1} thì y \(\in\) {-4; -2; 4; -10}

Mình làm mẫu phần này thì (x + 3)(y + 5) = -6 cũng vậy nha!

y + 3 chia hết cho x + 1 và 2x - 5 chia hết cho x + 4 mk làm sau nha!

Chúc bn học tốt

\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=x+1-\left(x-9\right)\)

\(\Rightarrow6x^2+27x+4x+18-6x^2-x-12x-2=10\)

\(\Rightarrow18x+16=10\)

\(\Rightarrow18x=-6\)

\(\Rightarrow x=-\frac{6}{18}=-\frac{1}{3}\)

cam on 

a,Để \(|2x+1|+|x-2|=0\Leftrightarrow\hept{\begin{cases}2x+1=0\\x-2=0\end{cases}}\)(vô lý)

=> ko có x thỏa mãn

b,\(|x+5|=2x-1\Leftrightarrow1-2x< x+5< 2x-1\)

giai gium mik cau c 

...... mk ko bít làm

\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

ta thấy \(2x^2+7>0\)

\(=>-3x-9< 0\)

\(=>-3x< 9\)

\(=>x>-3\)

vậy...