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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+3\right)}=\frac{20}{41}\)
\(2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}\right)=2.\frac{20}{41}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)
\(1-\frac{1}{x+2}=\frac{40}{41}\)
\(\frac{1}{x+2}=1-\frac{40}{41}\)
\(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 41 - 2
=> x = 39
Vẫy x = 39
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
=> \(1-\frac{1}{x+2}=\frac{40}{41}\)
=> \(\frac{1}{x+2}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
\(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
\(\Rightarrow\frac{27}{23}-\frac{4}{23}+\frac{1}{2}-\frac{4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
\(\Rightarrow1+0< x< \frac{7}{3}+\frac{3}{3}\)
\(\Rightarrow1< x< \frac{10}{3}\)
\(\Rightarrow1< x< 3,333333333\)
\(\Rightarrow x\in\left\{2;3\right\}\)
Vậy : ....
ta co : \(\frac{27}{23}+\frac{-4}{23}+\frac{1}{2}+\frac{-4}{8}< x< \frac{7}{3}+\frac{13}{41}+\frac{28}{41}\)
=> \(1< x< \frac{10}{3}\)
vi x la so nguyen => \(1< x\le3\)
con lai ban tu lam
Bài làm
\(\frac{15}{41}+\frac{-138}{41}\le x< \frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)
\(\frac{123}{41}\le x< 1\)
\(\frac{123}{41}\le x< \frac{41}{41}\)
\(\Rightarrow123\le x< 41\)
\(\Rightarrow x\in\varnothing\)
=> -123 / 41 < hoặc = x < 1
=> -3 < hoặc = x <1
=>x = ( -3 ; -2 ; -1 ; 0 )
2/1.3+2/3.5+...+2/x(x+2)= 40/41
1-1/3+1/3-1/5+...+1/x-1/(x+2)=40/41
1-1/(x+2)=40/41
1/(x+2)=1-40/41=1/41
x+2=41
x=41-2=39
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x\left(x+2\right)}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\frac{1}{2}.\frac{x+1}{x+2}=\frac{20}{41}\)
\(\frac{x+1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\frac{x+1}{x+2}=\frac{40}{41}\)
\(x+1=40
\)
\(x=40-1\)
\(x=39\)
Đúng thì ****
\(\left(x+1\right).\left(y+7\right)=41\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=41\\y+7=41\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=41-1\\y=41-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=40\\y=34\end{cases}}\)
Vậy .................
a: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
a: =>41-(2x-5)=720:40=18
=>2x-5=23
=>2x=28
=>x=14
b: =>100x+5050=5750
=>100x=700
=>x=7
Ta có: \(\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{x\left(x+2\right)}=\dfrac{20}{41}\)
\(\Leftrightarrow\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{40}{41}\)
\(\Leftrightarrow1-\dfrac{2}{x+2}=\dfrac{40}{41}\)
\(\Leftrightarrow\dfrac{2}{x+2}=\dfrac{1}{41}\)
Suy ra: x+2=82
hay x=80
a. 4.(x+41) = 7
x + 41 = 7 : 4 = 1,75
x = 1,75 - 41 = -39,25
b. 4.(x-3) = 72 - 110 = 49 - 1 = 48
x - 3 = 48 : 4 = 12
x = 12 + 3 = 15