\(\Leftrightarrow\dfrac{4\cdot90\cdot\left(x+5\right)-4\cdot90\cdot x}{4x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{4x\left(x+5\right)}\)

\(\Leftrightarrow x^2+5x-1800=0\)

\(\text{Δ}=5^2-4\cdot1\cdot\left(-1800\right)=7225>0\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-85}{2}=\dfrac{-90}{2}=-45\left(nhận\right)\\x_2=\dfrac{-5+85}{2}=40\left(nhận\right)\end{matrix}\right.\)

`(x/(x+1))^2+(x/(x-1))^2=90(x ne -1,1)`

`<=>x^2/(x+1)^2+x^2/(x-1)^2=90`

`<=>x^2(x-1)^2+x^2(x-1)^2=90(x^2-1)^2`

`<=>x^2(2x^2+2)=90(x^4-2x^2+1)`

`<=>2x^4+2x^2=90x^4-180x^2+90`

`<=>88x^4-182x^2+90=0`

`<=>88x^4-110x^2-72x^2+90=0`

`<=>22x^2(4x^2-5)-18(4x^2-5)=0`

`<=>(4x^2-5)(22x^2-18)=0`

`<=>(4x^2-5)(11x^2-9)=0`

`<=>` $\left[ \begin{array}{l}4x^2=5\\11x^2=9\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x=\sqrt{\dfrac{5}{4}}\\x=-\sqrt{\dfrac{5}{4}}\\x=\sqrt{\dfrac{9}{11}}\\x=-\sqrt{\dfrac{9}{11}}\end{array} \right.$

Vậy `S={\sqrt{9/11},-\sqrt{9/11},\sqrt{5/4},-\sqrt{5/4}}`

\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2=90\)

\(\Leftrightarrow\dfrac{x^2}{\left(x+1\right)^2}+\dfrac{x^2}{\left(x-1\right)^2}=90\)

\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}+\dfrac{x^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=90\)

\(\Leftrightarrow\dfrac{x^2\left(x-1\right)^2+x^2\left(x+1\right)^2-90\left(x-1\right)^2\left(x+1\right)^2}{\left(x+1\right)^2\left(x-1\right)^2}=0\)

\(\Rightarrow x^2\left(x^2-2x+1\right)+x^2\left(x^2+2x+1\right)-90\left(x^2-1\right)^2=0\)

\(\Leftrightarrow x^4-2x^3+x^2+x^4+2x^3+x^2-90x^4+90x^2-90=0\)

\(\Leftrightarrow-88x^4+92x^2-90=0\)

\(\left(\dfrac{x}{x+1}\right)^2+\left(\dfrac{x}{x-1}\right)^2+\dfrac{2x^2}{x^2-1}-\dfrac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\dfrac{x}{x+1}+\dfrac{x}{x-1}\right)^2-\dfrac{2x^2}{x^2-1}=90\)

\(\Leftrightarrow\left(\dfrac{2x^2}{x^2-1}\right)^2-\dfrac{2x^2}{x^2-1}-90=0\)

Đặt \(\dfrac{2x^2}{x^2-1}=t\Rightarrow t^2-t-90=0\Rightarrow\left[{}\begin{matrix}t=10\\t=-9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2x^2}{x^2-1}=10\\\dfrac{2x^2}{x^2-1}=-9\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}4x^2=5\\11x^2=9\end{matrix}\right.\)

\(\Leftrightarrow...\)

=>x+y=75 và 90x-90y=xy

=>y=75-x và 90x-90(75-x)=x(75-x)

=>y=75-x và 90x-6750+90x=75x-x^2

=>y=75-x và 180x-6750+x^2-75x=0

=>x^2+105x-6750=0 và y=75-x

=>x=45 hoặc x=-150 và y=75-x

=>\(\left(x,y\right)\in\left\{\left(45;30\right);\left(-150;225\right)\right\}\)

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

Em cảm ơn ạ.

Ta có:

\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)

\(\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)-x=\dfrac{-19}{24}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\dfrac{7}{30}-x=\dfrac{-19}{24}\)

\(x=\dfrac{7}{30}-\dfrac{-19}{24}\)

\(x=\dfrac{41}{40}\)

\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\dfrac{7}{30}-x=\dfrac{-19}{24}\)

\(\Rightarrow x=\dfrac{7}{30}-\dfrac{-19}{24}\)

\(\Rightarrow x=\dfrac{41}{40}\)

( 1/3.4 +1/4.5+ 1/5.6+ 1/6.7+1/7.8+1/8.9+1/9.10)-x=-19/24

(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)-x=-19/24

(1/3-1/10)-x=-19/24

7/30-x=-19/24

x=7/30-(-19)/24

x=41/40

nếu dung tick cho mình nhá

Mẫu số chung : \(LCM\left(60;120;36;90;72\right)=360\)

Quy đồng mẫu số :

\(\dfrac{360}{360}+\dfrac{-6}{360}+\dfrac{57}{360}< \dfrac{10\cdot x}{360}< \dfrac{232}{360}+\dfrac{295}{360}+\dfrac{-6}{360}\)

\(\Leftrightarrow\dfrac{411}{360}< \dfrac{10\cdot x}{360}< \dfrac{521}{360}\)

Vậy tập hợp các giá trị của x là \(x=\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)

\(1,\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y+4z}{2-18+12}=\dfrac{24}{-4}=-6\\ \Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-36\\z=-18\end{matrix}\right.\\ 2,\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50-34}{8}=\dfrac{16}{8}=2\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

\(3,6x=10y=15z\Leftrightarrow\dfrac{6x}{30}=\dfrac{10y}{30}=\dfrac{15z}{30}\\ \Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}=\dfrac{x+y-z}{5+3-2}=\dfrac{90}{6}=15\\ \Leftrightarrow\left\{{}\begin{matrix}x=75\\y=45\\z=30\end{matrix}\right.\)