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8 tháng 7 2023

,làm ơn giúp mik với ah

 

8 tháng 7 2023

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

\(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

\(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

\(\dfrac{337.2023}{2}\)

\(\dfrac{\text{681751}}{2}\)

c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)

\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)

\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)

\(=\dfrac{44}{75}\)

21 tháng 8 2021

Cảm ơn bạn lần nữa! vui

9 tháng 10 2019

\(a,\frac{x+8}{3}+\frac{x+7}{2}=-\frac{x}{5}\)

\(\Leftrightarrow\frac{10\cdot\left(x+8\right)}{30}+\frac{15\left(x+7\right)}{30}=\frac{-6x}{30}\)

\(\rightarrow10x+80+15x+105=-6x\)

\(\Leftrightarrow31x+185=0\)

\(\Leftrightarrow x=-\frac{185}{31}\)

b,\(b,\frac{x-8}{3}+\frac{x-7}{4}=4+\frac{1-x}{5}\)

\(\Leftrightarrow\frac{20\left(x-8\right)}{60}+\frac{15\left(x-7\right)}{60}=\frac{240}{60}+\frac{12\left(1-x\right)}{60}\)

\(\rightarrow20x-160+15x-105=240+12-12x\)

\(\Leftrightarrow47x-517=0\)\(\Leftrightarrow x=11\)

22 tháng 7 2023

a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{1}{2}\)

c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)

 

1 tháng 4 2018

          (2 x - 3) - (x + 2) = ( x - 2)-3(x - 5)

\(\Leftrightarrow\)2x - 3 - x - 2 = x - 2 - 3x + 15

\(\Leftrightarrow\)x - 5 = 13 - 2x

\(\Leftrightarrow\)3x = 18

\(\Leftrightarrow\)x = 6

  Vậy x = 6 là giá trị cần tìm

8 tháng 10 2021

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

\(\left(x+4\right)\left(x+4\right)=100\)

\(\left(x+4\right)^2=10^2\)

\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

8 tháng 10 2021

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

19 tháng 10 2019

a, |x-2|+x

TH1: |x-2|=x-2

=> |x-2|+x=x-2+x=2x-2 

TH2: |x-2|=-(x-2)= -x+2

=> |x-2|+x= -x+2+x=2

11 tháng 8 2019

\(\frac{3x}{10}+\frac{5}{2}+\frac{3}{5}=\frac{5}{2}\)

=> \(\frac{3}{10}x=-\frac{3}{5}\)

=>x=-2

11 tháng 8 2019

\(\left[30\%x+2+\frac{1}{2}\right]+\frac{3}{5}=2+\frac{1}{2}\)

\(\left[30\%x+\frac{5}{2}\right]+\frac{3}{5}=\frac{5}{2}\)

\(30\%+\frac{5}{2}=\frac{5}{2}-\frac{3}{5}\)

\(30\%x+\frac{5}{2}=\frac{19}{10}\)

\(\frac{3}{10}x=\frac{19}{10}-\frac{5}{2}\)

\(\frac{3}{10}x=\frac{-3}{5}\)

\(x=\frac{-3}{5}:\frac{3}{10}\)

\(x=-2\)

Vậy x=-2

15 tháng 8 2021

| x-2/3| = 1/3+2/5

             = 11/15

=> x-2/3=11/15          hoặc x-2/3=-11/15

            x= 7/5                      x       = -1/15

k cho mk nha

\(\left|x-\frac{2}{3}\right|-\frac{2}{5}=\frac{1}{3}\)

\(\Rightarrow\left|x-\frac{2}{3}\right|=\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\)

\(\Rightarrow\orbr{\begin{cases}\Rightarrow x-\frac{2}{3}=\frac{11}{15}\\x-\frac{2}{3}=\frac{-11}{15}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\Rightarrow x=\frac{7}{15}\\x=\frac{-1}{15}\end{cases}}\)