a) Ta có: \(\frac{x}{9}=\frac{-12}{27}\)

=> \(27.x=-12.9\)

=> \(27x=-108\)

=> \(x=108:27\)

=>\(x=4\)

Đặt \(A=\frac{2^{19}\cdot27^3+15\cdot4^9\cdot9^4}{6^9\cdot2^{10}+12^{10}}\)

\(A=\frac{2^{19}\cdot\left(3^3\right)^3+15\cdot\left(2^2\right)^9\cdot\left(3^2\right)^4}{6^9\cdot2^9\cdot2+12^{10}}\)

\(A=\frac{2^{19}\cdot3^9+15\cdot2^{18}\cdot3^8}{12^9\cdot2+12^9\cdot12}=\frac{\left(2^{18}\cdot3^8\right)\cdot6+\left(2^{18}\cdot3^8\right)\cdot15}{12^9\cdot\left(2+12\right)}\)

\(A=\frac{\left(2^{18}\cdot3^8\right)\cdot\left(6+15\right)}{12^9\cdot14}=\frac{2^{18}\cdot3^8\cdot21}{12^9\cdot14}=\frac{2^{18}\cdot3^8\cdot7\cdot3}{2^{18}\cdot3^9\cdot7\cdot2}=\frac{3^8\cdot3}{3^8\cdot3\cdot2}\)

\(A=\frac{1}{2}\)

Đặt \(B=\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}+\frac{4}{195}=\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+\frac{4}{11\cdot13}+\frac{4}{13\cdot15}\)

\(B=\frac{1}{2}\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{13}-\frac{4}{15}\right)\)

\(B=\frac{1}{2}\left(\frac{4}{5}-\frac{4}{15}\right)\)mà \(\frac{4}{5}-\frac{4}{15}< 1\Leftrightarrow\frac{1}{2}\left(\frac{4}{5}-\frac{4}{15}\right)< \frac{1}{2}\Leftrightarrow B< A\)

a) Ta có: \(\frac{x}{12}=\frac{y}{3}.\)

=> \(\frac{x}{12}=\frac{y}{3}\) và \(x-y=36.\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4.\)

\(\left\{{}\begin{matrix}\frac{x}{12}=4=>x=4.12=48\\\frac{y}{3}=4=>y=4.3=12\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(48;12\right).\)

b)

\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)

⇒ \(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)

⇒ \(\frac{5}{3}x=\frac{1}{21}\)

⇒ \(x=\frac{1}{21}:\frac{5}{3}\)

⇒ \(x=\frac{1}{35}\)

Vậy \(x=\frac{1}{35}.\)

\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

⇒ \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

⇒ \(x-\frac{1}{2}=\frac{1}{3}\)

⇒ \(x=\frac{1}{3}+\frac{1}{2}\)

⇒ \(x=\frac{5}{6}\)

Vậy \(x=\frac{5}{6}.\)

Có 1 câu bạn đăng mình làm ở dưới rồi mà.

Chúc bạn học tốt!

a)áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x}{12}=\frac{y}{3}=\frac{x-y}{12-3}=\frac{36}{9}=4\)

\(\)x/12=4 suy ra x=12.4=48

y/3=4 suy ra y=3.4 =12

b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)

\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)

\(\frac{5}{3}x=\frac{1}{21}\)

\(x=\frac{1}{21}:\frac{5}{3}\)

\(x=\frac{1}{35}\)

\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{2}{5}\)

\(x=\frac{-3}{20}\)

\(\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}\)

\(\left|x-\frac{2}{5}\right|=\frac{11}{4}-\frac{3}{4}\)

\(\left|x-\frac{2}{5}\right|=2\)

suy ra x-2/5=2 hoac x-2/5=-2

\(x-\frac{2}{5}=2\)

\(x=\frac{12}{5}\)

\(x-\frac{2}{5}=-2\)

\(x=\frac{-8}{5}\)

\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)

\(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)

\(x-\frac{1}{2}=\frac{1}{3}\)

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

\(x=\frac{5}{6}\)

a)tìm x,y biết:$\frac{x}{12}=\frac{y}{3}vàx-y=36$

Ta có : \(\dfrac{x}{12} = \dfrac{y}{3}\) \(\Rightarrow\) \(\dfrac{x-y}{12-3} = \dfrac{36}{9} = 4\)

b)tìm x biết:

$\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}$

\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{5}{7} - \dfrac{2}{3}\)

\(\Leftrightarrow\) \(\dfrac{5}{3}x = \dfrac{1}{21}\)

\(\Leftrightarrow\) \(x = \dfrac{1}{35}\)

$\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}$

\(\Leftrightarrow\) \(\dfrac{2}{5} +x\) = \(\dfrac{2}{3} - \dfrac{11}{12}\)

\(\Leftrightarrow\) \(\dfrac{2}{5} + x = \dfrac{-1}{4}\)

\(\Leftrightarrow\) \(x = \dfrac{-13}{20}\)

$\left|x-\frac{2}{5}\right|+\frac{3}{4}=\frac{11}{4}$

\(\Leftrightarrow\) \(x - \dfrac{2}{5} = 2\)

\(\Leftrightarrow\) \(x = \dfrac{8}{5}\)

${\left(x-\frac{1}{2}\right)}^3=\frac{1}{27}$

\(\Leftrightarrow\) \(x - \dfrac{1}{2} = \dfrac{1}{19683}\)

\(\Leftrightarrow\) \(x = \dfrac{19685}{39366}\)

Ta có: \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}\)

           \(\frac{-12}{-37}=\frac{12}{37}\frac{12}{37}hay\frac{13}{38}>\frac{-12}{-37}\)

\(\text{Ta thấy : }\frac{13}{38}>\frac{13}{39}=\frac{1}{3}\)

                     \(\frac{-12}{-37}\frac{-12}{-37}\)

Bài 1 

A= \(\frac{81^{10}.3^{17}}{27^{10}.9^{13}}\)

= \(\frac{\left(3^4\right)^{10}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{13}}\)

= \(\frac{3^{40}.3^{17}}{3^{30}.3^{26}}\)

= \(\frac{3^{57}}{3^{56}}\)= 3

\(\frac{x}{7}\)= \(\frac{-12}{49}\)

=> 49x = (-12) x 7 

=> 49x = -84

=> x= \(\frac{-12}{7}\)

1) \(A=\frac{3^{40}.3^{17}}{3^{30}.3^{26}}=\frac{3^{57}}{3^{56}}=3\)

2) \(\Leftrightarrow49x=-84\Rightarrow x=-\frac{12}{7}\)

k mik nha bn, tks

1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)

   \(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)

 ( . là nhân nha) 

    \(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)

   \(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)

\(x=\frac{113}{8}\)

( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)

   \(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{49}{81}=\frac{56}{81}\)

\(4y=\frac{7}{81}\)

y      =  7/81:4

y       = 7/324

a, (x³.x².x).(1/8-3/2+3/2-1)                      b, (x³.x²y.xy².y³).(1/8-1/4+1/12-1/27)

  =x⁶.(-7/8)                                                 = x⁶.y⁶.9.(-17/216)