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22 tháng 4 2022

Ta có:

    \(\dfrac{20}{x-3}\) = \(\dfrac{18}{x+2}\)  (x ≠ 3 và x ≠ -2)

⇔ \(\dfrac{20.\left(x+2\right)}{\left(x-3\right).\left(x+2\right)}\) = \(\dfrac{18.\left(x-3\right)}{\left(x+2\right).\left(x-3\right)}\)    

⇔ 20.(x + 2) = 18.(x - 3)

⇔  20x + 40 = 18x - 54

⇔ 18x - 20x = 40 + 54

⇔ -2x = 94

  ⇒   x = 94/-2

  ⇒   x = -47

 

8 tháng 6 2021

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

\(x=10\)

 

10 tháng 4 2021

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!

4 tháng 3 2022

\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)

\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)

a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)

hay x=1

b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)

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

d: \(\Leftrightarrow-8< x< 3\)

hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)

e: =>2/7-x=2/5

=>7-x=5

=>x=2

f: =>2x+3/3=10/3

=>2x+3=10

=>2x=7

=>x=7/2

g: =>(14+x)/7=15/7

=>x+14=15

=>x=1

h: =>(2x+3)/x=13/x

=>2x+3=13

=>2x=10

=>x=5

28 tháng 12 2021

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)

\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)

\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)

\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)

mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)

nên x-17=0

hay x=17

Vậy: x=17

b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)

\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)

\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)

mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)

nên x+20=0

hay x=-20

Vậy: x=-20

11 tháng 3 2022

a, \(\left(\dfrac{7}{2}-2x\right).\dfrac{10}{3}=\dfrac{22}{3}\Leftrightarrow\dfrac{7}{2}-2x=\dfrac{22}{10}=\dfrac{11}{5}\)

\(\Leftrightarrow2x=\dfrac{13}{10}\Leftrightarrow x=\dfrac{13}{20}\)

b, \(\dfrac{4x}{9}=\dfrac{9}{8}-\dfrac{125}{1000}=1\Leftrightarrow x=\dfrac{9}{4}\)

c, \(-\dfrac{x}{21}=\dfrac{60}{21}\Rightarrow x=-60\)

\(\Leftrightarrow\dfrac{x}{-2}=\dfrac{1}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}\)

=>x=1; y=36; z=12