\(\dfrac{x}{50}-\dfrac{x+12}{55}=1\)
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TD
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 5 2023
\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 4 = 0
(1 + \(\dfrac{55-x}{1963}\) ) + ( 1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0
\(\dfrac{1963+55-x}{1963}\) + \(\dfrac{1968+50-x}{1968}\)+\(\dfrac{1973+45-x}{1973}\)+\(\dfrac{1978+40-x}{1978}\)=0
\(\dfrac{2018-x}{1963}\)+\(\dfrac{2018-x}{1968}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1978}\)=0
(2018 - \(x\))\(\times\)( \(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0
2018 \(-x\) = 0
\(x\) = 2018
DN
17 tháng 5 2023
\(\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0\)
\(\Rightarrow\text{ }\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0\)
\(\Rightarrow\text{ }\left(\dfrac{55-x}{1963}+1\right)+\left(\dfrac{50-x}{1968}+1\right)+\left(\dfrac{45-x}{1973}+1\right)+\left(\dfrac{40-x}{1978}+1\right)=0\)
\(\Rightarrow\text{ }\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0\)
\(\Rightarrow\text{ }\left(2018-x\right)\left(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\right)=0\)
Mà \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\)
\(\Rightarrow\text{ }2018-x=0\)
\(\Rightarrow\text{ }x=2018-0\)
\(\Rightarrow\text{ }x=2018\)
Vậy, \(x=2018.\)
25 tháng 3 2022
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
18 tháng 3 2023
a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)
=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11
=>32/x=1/3-1/11=8/33
=>x=32:8/33=132
b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)
=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16
=>x=90
c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)
=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10
=>22/x=1/10*11/2=11/20=22/40
=>x=40
5 tháng 3 2022
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
DT
5 tháng 3 2022
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
BC
16 tháng 3 2022
Bài 3
\(\dfrac{55}{23}+\dfrac{-22}{23}\le x\le\dfrac{1}{5}-\dfrac{-1}{6}+\dfrac{79}{30}\)
\(=\dfrac{33}{23}\)\(\le x\le\dfrac{90}{30}\)
\(=\dfrac{33}{23}\le x\le3\)
Mà \(x\in Z\) \(\Rightarrow\)\(x=2\)
Có 1 giá trị thỏa mãn
Chọn A
Bài 4
\(\dfrac{-11}{12}< \dfrac{5}{x}< \dfrac{-11}{15}\)
Chọn D
Bài 5
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{100}{100}-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
CHọn C
27 tháng 4 2018
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
26 tháng 6 2017
a, \(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{x+1}{x-2}-\dfrac{x+7}{x+2}\right)=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{\left(x+1\right)\left(x+2\right)-\left(x-2\right)\left(x+7\right)}{\left(x-2\right)\left(x+2\right)}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left[\dfrac{x^2+3x+2-x^2-5x+14}{x^2-4}\right]=0\)
\(\Leftrightarrow\dfrac{12}{x^2-4}-\left(\dfrac{14-2x}{x^2-4}\right)=0\)
\(\Leftrightarrow12=14-2x\)
\(\Leftrightarrow x=1\)
Vậy x = 1
\(\Leftrightarrow\dfrac{x}{50}-\dfrac{x}{55}-\dfrac{12}{55}=1\)
\(\Leftrightarrow x\left(\dfrac{1}{50}-\dfrac{1}{55}\right)=1+\dfrac{12}{55}\)
\(\Leftrightarrow\dfrac{x}{550}=\dfrac{67}{55}\)
\(\Leftrightarrow x=670\)