( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

  ( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

   (\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

    (\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\)  = \(\dfrac{1}{18}\)

     ( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

       (\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)

         \(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)

               \(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)

               \(x\) = 4

x = 4 nhé

 ( 215152​ + 235352​ + 263632​) : �x = 118181​

  ( 2×715×715×72×7​ + 2×335×335×32×3​ + 263632​) : �x = 118181​

   (1410510514​ + 61051056​ + 263632​) : �x = 118181​

    (2010510520​ + 263632​) : �x  = 118181​

     ( 421214​ + 263632​) : �x = 118181​

       (12636312​ + 263632​) : �x = 118181​

         2992​ : �x = 118181​

               �x = 2992​ : 118181​

               �x = 4

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Cầu cứu mọi người hãy giúp đỡ mình

=>x(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9)+99/x=-3/7

=>8/9x+99/x=-3/7

\(\Leftrightarrow\dfrac{8x}{9}+\dfrac{99}{x}=\dfrac{-3}{7}\)

\(\Leftrightarrow\dfrac{8x^2+99\cdot9}{9x}=\dfrac{-3}{7}\)

\(\Leftrightarrow-56x^2-6237=27x\)

hay \(x\in\varnothing\)

b) \(\left(x-3\right)^2+3x-22=\sqrt{x^2-3x+7}\)

\(\Leftrightarrow x^2-6x+9+3x-22=\sqrt{x^2-3x+7}\)

\(\Leftrightarrow\left(x^2-3x+7\right)-\sqrt{x^2-3x+7}-20=0\)

Đặt \(\sqrt{x^2-3x+7}=t\left(t\ge0\right)\left(1\right)\)

\(\Rightarrow t^2-t-20=0\)

\(\Rightarrow x_1=5\left(TM\right);x_2=-4\left(KTM\right)\)

Thay t=5 vào (1), ta có :

\(\sqrt{x^2-3x+7}=5\)

\(\Leftrightarrow x^2-3x+7=25\)

\(\Leftrightarrow x^2-3x-18=0\)

\(\Rightarrow x_1=6;x_2=-3\)

vậy...

xl bn tớ gửi nhầm

\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+9\right)}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

=>\(\dfrac{x+9-x-1}{\left(x+9\right)\left(x+1\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow2\left(x^2+10x+9\right)=5\cdot8=40\)

=>x^2+10x+9=20

=>x^2+10x-11=0

=>(x+10)(x-1)=0

=>x=1 hoặc x=-10

\(\Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{2}{\left(x+3\right)\left(x+5\right)}+\dfrac{2}{\left(x+5\right)\left(x+7\right)}+\dfrac{2}{\left(x+7\right)\left(x+9\right)}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+9}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{x+9-x-1}{\left(x+1\right)\left(x+9\right)}=\dfrac{2}{5}\)

=>2(x+1)(x+9)=5*8=40

=>x^2+9x+9=20

=>x^2+9x-11=0

hay \(x=\dfrac{-9\pm5\sqrt{5}}{2}\)

=>x^2+9x

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

\(4.3^x+3^{x+1}=63\)

\(\Rightarrow4.3^x+3.3^x=63\)

\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)

\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)

mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)

\(\Rightarrow x\in\varnothing\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

a: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{24}{x-3}-\dfrac{10}{y+2}=126\\\dfrac{24}{x-3}+\dfrac{45}{y+2}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-55}{y+2}=165\\\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+2=\dfrac{-1}{3}\\\dfrac{12}{x-3}=48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)