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2 tháng 11 2021

\(20-2\left(x-1\right)^2=2\)

\(2\left(x-1\right)^2=20-2\)

\(2\left(x-1\right)^2=18\)

\(\left(x-1\right)^2=19:2\)

\(\left(x-1\right)^2=9\)

\(\left(x-1\right)^2=\pm3^2\)

\(\Rightarrow\hept{\begin{cases}x-1=3\\x-1=-3\end{cases}\Rightarrow\hept{\begin{cases}x=4\\x=-2\end{cases}}}\)

Vậy x = 4 ;x = -2 

16 tháng 8 2018

a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)

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

\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)

c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)

\(\Rightarrow x=9\)

DD
10 tháng 1 2021

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{15}\)

20 tháng 6 2019

\(a,\frac{1}{2}x+\frac{3}{5}\left(x-2\right)=3\)

\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{5}\right)=3+\frac{6}{5}\)

\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)

\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}\)

\(\Rightarrow x=\frac{42}{11}\)

20 tháng 6 2019

\(b,\left(3\frac{1}{2}-x\right).1\frac{1}{4}=-1\frac{1}{20}\)

\(\Rightarrow\left(\frac{7}{2}-x\right)\cdot\frac{5}{4}=-\frac{21}{20}\)

\(\Rightarrow\frac{7}{2}-x=-\frac{21}{20}:\frac{5}{4}\)

\(\Rightarrow\frac{7}{2}-x=-\frac{21}{25}\)

\(\Rightarrow x=\frac{7}{2}--\frac{21}{25}\)

\(\Rightarrow x=\frac{7}{2}+\frac{21}{25}\)

\(\Rightarrow x=\frac{217}{50}\)

5 tháng 11 2021

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}\)

=\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x+1\right)}\)

=\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)

=\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)\)

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

mình tháy bạn ko đề kết quả của phép tính nên chỉ lm được đến đây thôi

6 tháng 8 2016

\(a,\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\Leftrightarrow x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=1\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(\Leftrightarrow4\times x+\frac{15}{16}=1\)

\(\Leftrightarrow4\times x=1-\frac{15}{16}\)

\(\Leftrightarrow4\times x=\frac{1}{16}\)

\(\Leftrightarrow x=\frac{1}{16}\div4\)

\(\Leftrightarrow x=\frac{1}{64}\)

\(b,x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(\Leftrightarrow x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{13}+...+\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\left(\frac{1}{11}-\frac{1}{55}\right)\right]=\frac{3}{11}\)

\(\Leftrightarrow x-\left[715\times\frac{4}{55}\right]=\frac{3}{11}\)

\(\Leftrightarrow x-52=\frac{3}{11}\)

\(\Leftrightarrow x=\frac{3}{11}+52\)

\(\Leftrightarrow x=\frac{575}{11}\)

23 tháng 4 2019

a,x- 2/3= -1/2

x= -1/2 + 2/3

x= 1/6

b,20%.x - x +1/5 =3/4

1/5.x - x +1/5 = 3/4

1/5.x-x =3/4 - 1/5

x.(1/5 - 1)=11/20

x.-4/5 =11/20

x=11/20 : -4/5=-11/16

23 tháng 4 2019

a)  \(x-\frac{2}{3}=-\frac{1}{2}\)

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

\(\Leftrightarrow x=\frac{1}{6}\)

b)  \(20\%x-x+\frac{1}{5}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{5}x-x=\frac{3}{4}-\frac{1}{5}\)

\(\Leftrightarrow-\frac{4}{5}x=\frac{11}{5}\)

\(\Leftrightarrow x=\frac{11}{5}:\left(-\frac{4}{5}\right)\)

\(\Leftrightarrow x=-\frac{11}{4}\)

10 tháng 7 2017

Bài 3 : 

b) Ta có 1+ 2 + 3 +4 + ...+ x =15

Nên \(\frac{x\left(x+1\right)}{2}=15\)

\(x\left(x+1\right)=30\)

=> \(x\left(x+1\right)=5.6\)

=> x = 5

19 tháng 6

Bài 2:

h; \(\dfrac{2}{3}\)\(x\)  + 50% + \(x\) = \(\dfrac{1}{10}\)

    \(\dfrac{2}{3}\)\(x\) + \(\dfrac{1}{2}\)  + \(x\) = \(\dfrac{1}{10}\)

    (\(\dfrac{2}{3}\)\(x\) + \(x\)) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

     \(x\) \(\times\) (\(\dfrac{2}{3}\) + 1) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) + \(\dfrac{1}{2}\) = \(\dfrac{1}{10}\)

      \(x\) \(\times\) \(\dfrac{5}{3}\) = \(\dfrac{1}{10}\) - \(\dfrac{1}{2}\)

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

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

      \(x\)         =   - \(\dfrac{6}{25}\) 

Lớp 5 chưa học số âm em nhé. 

17 tháng 11 2023

a: x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b: 2x(x+3)=0

=>x(x+3)=0

=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

c: \(\left(6-x\right)\left(x+10\right)=0\)

=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)

d: \(\left(5x+20\right)\left(x^2+1\right)=0\)

=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)

=>5x=-20

=>x=-4