Ta có:

\(x^2+2x=y^2\\ \Leftrightarrow x^2+2x+1=y^2+1\\ \Leftrightarrow\left(x+1\right)^2-y^2=1\\ \Leftrightarrow\left(x+1-y\right)\left(x+1+y\right)=1=1.1\)

Vì x,y nguyên nên x+1-y và x+1+y nguyên. 

Do đó: \(\left\{{}\begin{matrix}x+1-y=1\\x+1+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-y=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Cặp số nguyên (x,y) thõa mãn bài toán là (0;0)

     ( 10 - \(\dfrac{5}{3}\) + \(\dfrac{4}{9}\)) - ( 7- \(\dfrac{8}{3}\)- \(\dfrac{4}{9}\))

=    10 - \(\dfrac{5}{3}\) + \(\dfrac{4}{9}\) - 7 + \(\dfrac{8}{3}\) + \(\dfrac{4}{9}\)

= ( 10 - 7) +   ( \(\dfrac{8}{3}\)- \(\dfrac{5}{3}\)) + ( \(\dfrac{4}{9}\) + \(\dfrac{4}{9}\))

= 3 + \(\dfrac{3}{3}\) + \(\dfrac{8}{9}\)

= 4 + \(\dfrac{8}{9}\)

= \(\dfrac{36}{9}\) + \(\dfrac{8}{9}\)

= \(\dfrac{44}{9}\)

TH1: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

\(\Rightarrow\left(2x-3\right)^2=3-2x\)

\(\Leftrightarrow\left(2x-3\right)^2+2x-3=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x-3+1\right)=0\)

Giải nhanh: \(x=\dfrac{3}{2}\)(nhận) hay \(x=1\) (nhận)

TH2: \(3-2x< 0\Leftrightarrow x>\dfrac{3}{2}\)

\(\Rightarrow\left(2x-3\right)^2=2x-3\)

Giải tương tự, ta suy ra \(x=\dfrac{3}{2}\) (loại) hay \(x=2\) (nhận)

Kết luận phương trình có 3 nghiệm \(x=1;x=\dfrac{3}{2};x=2\)

\(\dfrac{3}{5}.\dfrac{5}{7}+\dfrac{3}{5}:\dfrac{5}{3}-\dfrac{1}{7}:\dfrac{5}{3}\)

= \(\dfrac{3}{5}.\dfrac{5}{7}-\dfrac{3}{7}.\dfrac{3}{5}-\dfrac{1}{7}.\dfrac{3}{5}\)

=\(\dfrac{3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}-\dfrac{1}{7}\right)\)

=\(\dfrac{3}{5}.1\)

=\(\dfrac{3}{5}\)

a) A = $\frac{124}{125}+(\frac{-18}{15}+\frac{1}{125})+\frac{3}{15}$

     = $(\frac{124}{125}+\frac{1}{125})+(\frac{-18}{15}++\frac{3}{15})$

     = $\frac{124+1}{125}+\frac{-18+3}{15}$

     = 1 + (-1)

     = 0 
  

b) B = $(\frac{1}{9}-\frac{5}{17})+\frac{3}{6}+(\frac{-12}{17}+\frac{-1}{2})+\frac{5}{9}$

     = $(\frac{1}{9}+\frac{5}{9})+(\frac{-12}{17}-\frac{5}{17})+(\frac{3}{6}+\frac{-1}{2})$

     = $\frac{1+5}{9}+\frac{-12-5}{-17}+\frac{1+(-1)}{2}$

     = $\frac{2}{3}-1+0$

     = $\frac{-1}{\mathrm{3\; like\; đê}}$

`500 - {5.[409 - (2^3 . 3 - 21)^3 ] - 1724}`

`=500 - {5.[409 - (8.3-21)^3 ] - 1724}`

`=500- {5 . [409 - (24-21)^3 ] - 1724}`

`=500 - {5 .[409 - 3^3] - 1724}`

`=500 - {5.382 - 1724}`

`=500 - {1910 - 1724}`

`=500 - 186`

`=314`

_______________________________________

`375 : { 32 - [4 + ( 5.3^2 - 42)]} -14`

`=375 :{ 32 - [4 + ( 5 . 9 - 42)]} -14`

`=375 : { 32 - [4 + (45-42)]}-14`

`=375 :{32-[4+3]}-14`

`=375 : 25-14`

`=15-14`

`=1`