Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}=\frac{x-y+x+y}{16}=\frac{2x}{16}=\frac{x}{8}=\frac{25x}{200}=\frac{xy}{200}\)

Suy ra: \(25x=xy\Rightarrow y=25\)

Ta có: \(\frac{x-y}{3}=\frac{x+y}{13}\)

Suy ra: \(13x-13y=3x+3y\)

Thế y vào đẳng thức trên:

\(13x-325=3x+75\)

Suy ra: \(10x=325+75=400\Rightarrow x=40\)

Vậy ........

 (x-3)^2 *(x+3) *(x-3) =

 (x+4)^2 -(x-2)^2 -2(x+2) *(x-2) =

a) \(\left(x-3\right)^2.\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right).\left(x-3\right).\left(x+3\right).\left(x-3\right)\)

\(=\left(x-3\right)^3.\left(x+3\right)\)

\(=\left(3x-9\right).\left(x+3\right)\)

Phần b tương tự

X-3/5=1+2/3                                                4/7:X=1/2x2/5

X-3/5=5/3                                                    4/7:X=1/5

X=5/3+3/5                                                   X=4/7:1/5

X=34/15                                                      X=4

Vậy X=34/15                                          Vậy X=4

X - 3/5 = 1 + 2/3.         

<=> X = 1 + 2/3 + 3/5

<=> X =  15/15 + 10/15 + 9/15

<=> X = 34/15

4/7 : x = 1/2 x 2/5

<=> X = 4/7 : (1/2 x 2/5)

<=> X = 4/7 : 2/10

<=> X = 4/7 x 10/2

<=> X = 40/14

<=> X = 20/7

x- 3/5= 1+ 2/3                                     4/7 :x =1/2x 2/5

x- 3/5= 5/3                                          4/7: x = 1/5

x       = 5/3+ 3/5                                   x        = 4/7: 1/5

x      = 34/15                                        x        = 20/7

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)

\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)

\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)

\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)

\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)

\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\)         ( 4^3 = 8^2 ; 5^4 = 25^2 )

\(=\frac{1}{25\times7}=\frac{1}{175}\)

\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)

\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)

\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)

\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)

a) \(\frac{-9}{12}=\frac{-9\div3}{12\div3}=\frac{-3}{4}\)

\(\frac{-18}{-24}=\frac{-18\div\left(-6\right)}{-24\div\left(-6\right)}=\frac{3}{4}\)

\(\frac{-35}{70}=\frac{-35\div35}{70\div35}=\frac{-1}{2}\)

\(\frac{-9}{27}=\frac{-9\div9}{27\div9}=\frac{-1}{3}\)

b) \(\frac{1313}{4242}=\frac{1313\div101}{4242\div101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{-353535\div\left(-70707\right)}{-424242\div\left(-70707\right)}=\frac{5}{6}\)

c)\(\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{\left(2^3\right)^2\times\left(5^2\right)^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^6\times5^6\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^3\times2^3\times5^2\times5^4\times7}\)= \(\frac{2^3\times2^3}{2^3\times5^2\times7}\)= \(\frac{8}{25\times7}\)= \(\frac{8}{175}\)

a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)

\(\Leftrightarrow9x-14x-8=5\)

\(\Leftrightarrow-5x-8=5\)

\(\Leftrightarrow-5x=5+8\)

\(\Leftrightarrow-5x=13\)

\(\Rightarrow x=-\dfrac{13}{5}\)

Vậy \(x=-\dfrac{13}{5}\)

b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)

\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)

đến đây bạn giải tiếp nhé

c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)