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`a)` Xét tử số phân số M :
\(2012-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{2012}{2020}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{2012}{2020}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{2020}\\ =24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)\)
Ta được : \(M=\dfrac{24\left(\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}\right)}{\dfrac{1}{27}+\dfrac{1}{30}+\dfrac{1}{33}+...+\dfrac{1}{6060}}=24\)
`b)` Xét tử số phân số N :
\(\dfrac{1}{1.300}+\dfrac{1}{2.301}+\dfrac{1}{3.302}+...+\dfrac{1}{101.400}\\ =\dfrac{1}{299}.\left(\dfrac{299}{1.300}+\dfrac{299}{2.301}+\dfrac{299}{3.302}+...+\dfrac{299}{101.400}\right)\\ =\dfrac{1}{299}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Xét mẫu số phân số N :
\(\dfrac{1}{1.102}+\dfrac{1}{2.103}+\dfrac{1}{3.104}+...+\dfrac{1}{299.400}\\ =\dfrac{1}{101}.\left(\dfrac{101}{1.102}+\dfrac{101}{2.103}+\dfrac{101}{3.104}+...+\dfrac{101}{299.400}\right)\\ =\dfrac{1}{101}.\left(1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+\dfrac{1}{3}-\dfrac{1}{104}+...+\dfrac{1}{299}-\dfrac{1}{400}\right)\)
\(=\dfrac{1}{101}.\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)\)
Ta được: \(N=\dfrac{\dfrac{1}{299}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}{\dfrac{1}{101}\left(1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+\dfrac{1}{3}-\dfrac{1}{302}+...+\dfrac{1}{101}-\dfrac{1}{400}\right)}\\ =\dfrac{\dfrac{1}{299}}{\dfrac{1}{101}}=\dfrac{101}{299}\)
\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(=1-\dfrac{6}{21}=\dfrac{15}{21}=\dfrac{5}{7}\)
a) \(\dfrac{3}{4}+\dfrac{1}{4}\) : x \(=\) \(\dfrac{2}{5}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{2}{5}\) \(-\) \(\dfrac{3}{4}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{8}{20}\) \(-\) \(\dfrac{15}{20}\)
\(\dfrac{1}{4}\) : x \(=\) \(\dfrac{-7}{20}\)
x \(=\) \(\dfrac{1}{4}\) : \(\dfrac{-7}{20}\)
x \(=\) \(\dfrac{1}{4}.\dfrac{-20}{7}\)
x \(=\) \(\dfrac{-5}{7}\)
Vậy x \(=\) \(\dfrac{-5}{7}\)
b) \(\dfrac{1}{4}+\dfrac{1}{3}\) : 2x \(=\) \(-5\)
\(\dfrac{1}{3}:2\)x \(=\) \(-5-\dfrac{1}{4}\)
\(\dfrac{1}{3}:2x\) \(=\) \(\dfrac{-20}{4}-\dfrac{1}{4}\)
\(\dfrac{1}{3}:2x\) \(=\) \(\dfrac{-21}{4}\)
2x \(=\) \(\dfrac{1}{3}:\dfrac{-21}{4}\)
2x \(=\) \(\dfrac{1}{3}.\dfrac{-4}{21}\)
2x \(=\) \(\dfrac{-4}{63}\)
x \(=\) \(\dfrac{-4}{63}:2\)
x \(=\) \(\dfrac{-4}{63}.\dfrac{1}{2}\)
x \(=\) \(\dfrac{-2}{63}\)
Vậy x \(=\) \(\dfrac{-2}{63}\)
c) \(-2\) \(+\) \(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\) \(\dfrac{9}{10}\) \(-\left(-2\right)\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}+2\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{9}{10}+\dfrac{20}{10}\)
\(\left|x+\dfrac{1}{5}\right|\) \(=\dfrac{29}{10}\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{29}{10}\\x+\dfrac{1}{5}=\dfrac{-29}{10}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{29}{10}-\dfrac{1}{5}\\x=\dfrac{-29}{10}-\dfrac{1}{5}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{29}{10}-\dfrac{2}{10}\\x=\dfrac{-29}{10}-\dfrac{2}{10}\end{matrix}\right.\)
\(\rightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{27}{10}=2,7\\x=\dfrac{-31}{10}=-3,1\end{matrix}\right.\)
Vậy x \(=\) 2,7 hoặc \(-3,1\)
a) \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{-7}{20}\)
\(\Leftrightarrow x=\dfrac{1}{4}:\dfrac{-7}{20}\)
\(\Leftrightarrow x=-\dfrac{5}{7}\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}>\dfrac{1}{14}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}>0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Chúc bạn học tốt!!!
Giải:
a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Leftrightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Leftrightarrow x+1\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Dấu "=" xảy ra:
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}=0\end{matrix}\right.\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
b) \(\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(-\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\)
\(=\left(\dfrac{3}{9}-\dfrac{2}{9}\right)+\left(-\dfrac{27}{36}-\dfrac{1}{36}\right)+\left(\dfrac{9}{15}+\dfrac{1}{15}\right)+\dfrac{1}{64}\)
\(=\dfrac{1}{9}-\dfrac{7}{9}+\dfrac{2}{3}+\dfrac{1}{64}=0+\dfrac{1}{64}=\dfrac{1}{64}\)
Chúc bạn học tốt!
a: \(\Leftrightarrow x^2=900\)
=>x=30 hoặc x=-30
b: \(\Leftrightarrow\dfrac{2}{3}:\left(-0.1x\right)=\dfrac{4}{3}:\dfrac{-2}{25}=-\dfrac{4}{3}\cdot\dfrac{25}{2}=-\dfrac{100}{6}=\dfrac{-50}{3}\)
=>0,1x=2/3:50/3=2/3x3/50=1/25
=>1/10x=1/25
hay x=1/25:1/10=10/25=2/5
d: \(\Leftrightarrow x^2=\dfrac{144}{25}\)
=>x=12/5 hoặc x=-12/5
Câu 2:
(x-4/7)(x+1/2)>0
=>x-4/7>0 hoặc x+1/2<0
=>x>4/7 hoặc x<-1/2
Sửa đề: \(B=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2022\cdot2024}\right)\)
\(=\left(1+\dfrac{1}{2^2-1}\right)\cdot\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2023^2-1}\right)\)
\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2023^2}{2023^2-1}\)
\(=\dfrac{2\cdot3\cdot...\cdot2023}{1\cdot2\cdot...\cdot2022}\cdot\dfrac{2\cdot3\cdot....\cdot2023}{3\cdot4\cdot...\cdot2024}\)
\(=\dfrac{2023}{1}\cdot\dfrac{2}{2024}=\dfrac{2023}{1012}\)