\(\frac{x}{4}=\frac{18}{x+1}\)

\(\Leftrightarrow x\left(x+1\right)=72\)

\(\Leftrightarrow x=8\)

P/s tham khảo nha

bn trần hoàng việt ơi, bn có thể giải kĩ hơn đc chứ?

(suy ra )  x(x+1)=72

               x(x+1)=8(7+1)

        Vậy x=8 

mình đảm bảo đúng

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

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

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

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

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

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

\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\)         ( 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)\(\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{\left(2^3\right)^2\times\left(5^2\right)^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^6\times5^6\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^3\times2^3\times5^2\times5^4\times7}\)= \(\frac{2^3\times2^3}{2^3\times5^2\times7}\)= \(\frac{8}{25\times7}\)= \(\frac{8}{175}\)

y;x=(2;1),(-2;1),(-4;0)

\(\left(2x+1\right)\cdot\left(y-5\right)=12\)

<=>\(x=\frac{17-y}{2y-10}\)

thay x vào phương trình 

=>\(\left(\frac{17-y+y-5}{y-5}\right)\cdot\left(y-5\right)=12\)

<=>\(\frac{12}{y-5}\cdot\left(y-5\right)=12\)

<=>\(12=12\)(Luôn đúng khi và chỉ khi y khác 5 )\(y\ne5,y\inℝ\)

giả sử thay y=1 ta có 

=>\(2x=\frac{12}{1-5}-1\)

<=>\(2x=-4\)

=>\(x=-2\)

Vậy \(x=-2\)và \(y=1\)

\(\frac{x+1}{3}=\frac{9}{2}\)

\(\left(x+1\right).2=9.3\)

\(\left(x+1\right).2=27\)

\(x+1=27:2\)

\(x+1=13,5\)

\(x=13,5-1=12,5\)

vậy x = 12.5

\(\frac{x+1}{3}=\frac{9}{2}\)

\(\Leftrightarrow2\left(x+1\right)=3\times9\)

\(\Leftrightarrow2\left(x+1\right)=27\)

\(\Leftrightarrow x+1=\frac{27}{2}\)

\(\Leftrightarrow x=\frac{25}{2}\)

\(\frac{x+1}{3}\)= \(\frac{9}{2}\)

:> 2( x +1) = 9 . 3

:> 2x + 2 = 27

:>       2x = 27 - 2

:>       2x = 25

:>        x = \(\frac{25}{2}\)

Cái này chắc là đề THCS chứ không phải toán lớp 2 bạn à. !! >_<

(1+x).{(x-1):4+1}=501501

........tự giải tiếp