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Đề tớ gõ sai, Sr các cậu...
Đề đúng là :
\(\frac{x-3}{90}+\frac{x-2}{91}+\frac{x-1}{92}=3\)
Giúp tớ nhen...Giải chi tiết giùm nha...Thank you !!!
\(\left(\frac{x-3}{90}-1\right)+\left(\frac{x-2}{91}-1\right)+\left(\frac{x-1}{90}-1\right)=0\)
\(\Leftrightarrow\frac{x-93}{90}+\frac{x-93}{91}+\frac{x-93}{92}=0\)
\(\Leftrightarrow\left(x-93\right)\left(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\right)=0\)
mà \(\frac{1}{90}+\frac{1}{91}+\frac{1}{92}\ne0\)
\(\Leftrightarrow x-93=0\Leftrightarrow x=93\)
Vậy x=93
Gửi link thì bị lỗi, thôi nhai lại v:
Xét VT__Ta có: \(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{50}\right)\)
\(=\) \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{50}-1+\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
Ta có : \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}.....+\frac{1}{50}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-....-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+.......+\frac{1}{50}\)
a: Đề thiếu vế phải rồi bạn
b: \(\Leftrightarrow x\cdot\dfrac{-12}{13}=6+\dfrac{1}{13}+5=11+\dfrac{1}{13}=\dfrac{144}{13}\)
hay x=-12
c: \(\Leftrightarrow x\cdot\dfrac{-5}{4}-\dfrac{6}{5}x=\dfrac{-1}{2}-\dfrac{3}{7}\)
\(\Leftrightarrow x\cdot\dfrac{-49}{20}=\dfrac{-13}{14}\)
hay x=130/343
mk chỉ biết câu a thôi
\(-3+\frac{1}{3}\)=\(\frac{-3}{1}\)+\(\frac{1}{3}\)= \(\frac{-9}{3}\)+\(\frac{1}{3}\)=\(\frac{-8}{3}\)
Bài 1:
= \(\frac{-8}{3}\)
= \(\frac{-9}{4}\)
= \(\frac{-42}{19}\)
Bài 2:
A = \(\frac{3}{7}\)
Ai thấy mình đúng tk nha !
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x-1}{3}+\frac{x-1}{4}+\frac{x-1}{5}\)
\(\Leftrightarrow\frac{x-1}{1}+\frac{x-1}{2}-\frac{x-1}{3}-\frac{x-1}{4}-\frac{x-1}{5}=0\)
\(\Leftrightarrow\left(x-1\right)\left(1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\ne0\right)=0\)
\(\Leftrightarrow x=1\)
\(A=xemlai\) chưa hưa hiểu Quy luật
\(B=\frac{\left(n.\left(n+2\right)+1\right)}{n\left(n+2\right)}=\frac{\left(n+1\right)^2}{n.\left(n+2\right)}\)
\(B=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.5}...\frac{98.98}{97.99}\frac{99.99}{98.100}\frac{100.100}{99.101}\\\)
\(B=\frac{2.100}{1.101}=\frac{200}{101}\)
\(\frac{-1}{91}+\frac{-1}{247}+\frac{-1}{475}+\frac{-1}{775}+\frac{-1}{1147}\)
\(=-\left(\frac{1}{7.13}+\frac{1}{13.19}+\frac{1}{19.25}+\frac{1}{25.31}+\frac{1}{31.37}\right)\)
\(=-[\frac{1}{6}.\left(\frac{6}{7.13}+\frac{6}{13.19}+\frac{6}{19.25}+\frac{6}{25.31}+\frac{6}{31.37}\right)]\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+\frac{1}{19}-\frac{1}{25}+\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\left(\frac{1}{7}-\frac{1}{37}\right)\text{]}\)
\(=-\text{[}\frac{1}{6}.\frac{30}{259}\text{]}\)
\(=-\frac{5}{259}\)