X+1/10+x+1/11=x+1/12
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{9x10}\)\(+\frac{1}{10x11}\)\(+\frac{1}{11x12}\)\(+.....\)\(+\frac{1}{805x806}\)
\(=\frac{1}{9}\)\(-\frac{1}{10}\)\(+\frac{1}{10}\)\(-\frac{1}{11}\)\(+\frac{1}{11}\)\(-\frac{1}{12}\)\(+.....\frac{1}{805}\)\(-\frac{1}{806}\)
\(=\frac{1}{9}\)\(-\frac{1}{806}\)
\(=\frac{797}{7254}\)
a,\(\frac{11}{12}-\left(\frac{5}{42}-x\right)=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow\frac{11}{12}-\frac{5}{42}+x=\frac{15}{28}-\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{28}-\frac{11}{12}-\frac{11}{12}+\frac{5}{42}\)
\(\Leftrightarrow x=\left(\frac{15}{28}+\frac{5}{42}\right)-\left(\frac{11}{12}+\frac{11}{12}\right)\)
\(\Leftrightarrow x=\frac{55}{84}-\frac{11}{6}\)
\(\Leftrightarrow x=\frac{-33}{28}\)
b, \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Đặt:
\(X=\left(1+\dfrac{1}{9}\right)\left(1+\dfrac{1}{10}\right)\left(1+\dfrac{1}{11}\right).....\left(1+\dfrac{1}{200}\right)\)
\(X=\dfrac{10}{9}.\dfrac{11}{10}.\dfrac{12}{11}......\dfrac{201}{200}\)
\(X=\dfrac{10.11.12......201}{9.10.11......200}\)
\(X=\dfrac{201}{9}\)
\(Y=\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right).....\left(1-\dfrac{1}{99}\right)\)
\(Y=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}\)
\(Y=\dfrac{9.10.11......98}{10.11.12.....99}\)
\(Y=\dfrac{9}{99}=\dfrac{1}{11}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
=> x + 1 = 0 ( vì 1/10 + 1/11 + 1/12 - 1/13 - 1/14 khac 0 )
=> x = -1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
mà 1/10 > 1/13; 1/11>1/14
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)
=> x + 1 = 0
x = -1
x+110+x+111+x+112=x+113+x+114x+110+x+111+x+112=x+113+x+114
⇒x+110+x+111+x+112−x+113−x+114=0⇒x+110+x+111+x+112−x+113−x+114=0
⇒(x+1).(110+111+112−113−114)=0⇒(x+1).(110+111+112−113−114)=0
mà 1/10 > 1/13; 1/11>1/14
⇒110+111+112−113−114≠0⇒110+111+112−113−114≠0
=> x + 1 = 0
x = -1
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)
=>x+1=0
hay x=-1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow x+1=0\)( do \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\))
\(\Rightarrow x=-1\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)(1)
Rõ ràng: \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)>0\)nên (1) <=> x+1 = 0; hay x = -1.
\(x+\frac{1}{10}+x+\frac{1}{11}=x+\frac{1}{12}\)
\(\Leftrightarrow11\left(x+1\right)+10\left(x+1\right)=\frac{x+1}{12}\)
\(\Leftrightarrow12\left(11x+11+10x+10\right)=x+1\)
\(\Leftrightarrow132x+132+120x+120=x+1\)
\(\Leftrightarrow251x=-251\)
\(\Leftrightarrow x=-251:251\)
\(\Leftrightarrow x=-1\)
\(x+\frac{1}{10}+x+\frac{1}{11}=x+\frac{1}{12}\)
\(\Leftrightarrow2x-x=\frac{1}{12}-\frac{1}{10}-\frac{1}{11}\)
\(\Leftrightarrow x=-\frac{71}{660}\)