\(\left(100+\dfrac{99}{2}+\dfrac{98}{3}+\dfrac{97}{4}....+\dfrac{1}{100}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....\dfrac{1}{100}\right)-2\)
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23 tháng 1 2021
a) Ta có: \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90}{15}-\dfrac{5\left(1-2x\right)}{15}\)
\(\Leftrightarrow3x-9=90-5+10x\)
\(\Leftrightarrow3x-9=10x+85\)
\(\Leftrightarrow3x-10x=85+9\)
\(\Leftrightarrow-7x=94\)
hay \(x=-\dfrac{94}{7}\)
Vậy: \(S=\left\{-\dfrac{94}{7}\right\}\)
b) Ta có: \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\)
\(\Leftrightarrow\dfrac{2\left(3x-2\right)}{12}-\dfrac{60}{12}=\dfrac{3\left(3-2x-14\right)}{12}\)
\(\Leftrightarrow6x-4-60=9-6x-42\)
\(\Leftrightarrow6x-64=-6x-33\)
\(\Leftrightarrow6x+6x=-33+64\)
\(\Leftrightarrow12x=31\)
hay \(x=\dfrac{31}{12}\)
Vậy: \(S=\left\{\dfrac{31}{12}\right\}\)
c) Ta có: \(3\left(x-1\right)+3=5x\)
\(\Leftrightarrow3x-3+3=5x\)
\(\Leftrightarrow3x-5x=0\)
\(\Leftrightarrow-2x=0\)
hay x=0
Vậy: S={0}
d) Ta có: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
\(\Leftrightarrow\dfrac{x+1}{100}+1+\dfrac{x+2}{99}+1=\dfrac{x+3}{98}+1+\dfrac{x+4}{97}+1\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}=\dfrac{x+101}{98}+\dfrac{x+101}{97}\)
\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\)
\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\)
mà \(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\ne0\)
nên x+101=0
hay x=-101
Vậy: S={-101}
23 tháng 1 2021
a) \(\dfrac{x-3}{5}=6-\dfrac{1-2x}{3}\\ \Leftrightarrow\dfrac{3\left(x-3\right)}{15}=\dfrac{90-5\left(1-2x\right)}{15}\\ \Leftrightarrow3x-9=90-5+10x\\ \Leftrightarrow3x-10x=90-5+9\\ \Leftrightarrow-7x=94\\ \Leftrightarrow x=\dfrac{-94}{7}\)
Vậy \(x=\dfrac{-94}{7}\) là nghiệm của pt
b) \(\dfrac{3x-2}{6}-5=\dfrac{3-2\left(x+7\right)}{4}\\ \Leftrightarrow\dfrac{2\left(3x-2\right)-60}{12}=\dfrac{9-6\left(x+7\right)}{12}\\ \Leftrightarrow6x-4-60=9-6x-42\\ \Leftrightarrow6x+6x=9-42+4+60\\ \Leftrightarrow12x=31\\ \Leftrightarrow x=\dfrac{31}{12}\)
Vậy \(x=\dfrac{31}{12}\) là nghiệm của pt
c) \(3\left(x-1\right)+3=5x\\ \Leftrightarrow3x+3+3=5x\\ \Leftrightarrow5x-3x=3+3\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)
Vậy x = 3 là nghiệm của pt
d) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\\ \Leftrightarrow\left(\dfrac{x+1}{100}+1\right)+\left(\dfrac{x+2}{99}+1\right)=\left(\dfrac{x+3}{98}+1\right)+\left(\dfrac{x+4}{97}+1\right)\\ \Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}-\dfrac{x+101}{98}-\dfrac{x+101}{97}=0\\ \Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{98}-\dfrac{1}{97}\right)=0\\ \Leftrightarrow x+101=0\\ \Leftrightarrow x=-101\)
Vậy x = -101 là nghiệm của pt
e) \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}=-4\\ \Leftrightarrow\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{53-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)=0\\ \Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}\right)=0\\ \Leftrightarrow100-x=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
f) \(\dfrac{x-90}{10}+\dfrac{x-76}{12}+\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\\ \Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\\ \Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\\ \Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\\ \Leftrightarrow x-100=0\\ \Leftrightarrow x=100\)
Vậy x = 100 là nghiệm của pt
9 tháng 1
a: \(\dfrac{3}{4}+\dfrac{1}{4}:x=-2\dfrac{1}{2}\)
=>\(\dfrac{1}{4}:x=-\dfrac{5}{2}-\dfrac{3}{4}=-\dfrac{10}{4}-\dfrac{3}{4}=-\dfrac{13}{4}\)
=>\(x=\dfrac{-1}{4}:\dfrac{13}{4}=\dfrac{-1}{4}\cdot\dfrac{4}{13}=\dfrac{-1}{13}\)
b: \(\left(\dfrac{2}{3}\right)^{100}:x=\left(-\dfrac{2}{3}\right)^{98}\)
=>\(\left(\dfrac{2}{3}\right)^{100}:x=\left(\dfrac{2}{3}\right)^{98}\)
=>\(x=\left(\dfrac{2}{3}\right)^{100}:\left(\dfrac{2}{3}\right)^{98}=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
c: \(\dfrac{3}{2}:\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{4}\)
=>\(\left|4x-\dfrac{1}{5}\right|=\dfrac{3}{2}:\dfrac{3}{4}=\dfrac{3}{2}\cdot\dfrac{4}{3}=2\)
=>\(\left[{}\begin{matrix}4x-\dfrac{1}{5}=2\\4x-\dfrac{1}{5}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{5}\\4x=-\dfrac{9}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{11}{20}\\x=-\dfrac{9}{20}\end{matrix}\right.\)
19 tháng 6 2017
\(\dfrac{100+\dfrac{99}{2}+\dfrac{98}{3}+...+\dfrac{1}{100}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\left(\dfrac{99}{2}+1\right)+\left(\dfrac{98}{3}+1\right)+...+\left(\dfrac{1}{100}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{\dfrac{101}{2}+\dfrac{101}{3}+...+\dfrac{101}{100}+\dfrac{101}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=\dfrac{101\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}+\dfrac{1}{101}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{101}}-2\)
\(=101-2\)
\(=99\)
17 tháng 6 2018
a, \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ 3B=3+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ 3B-B=\left(3+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\2B=3-\dfrac{1}{3^{2005}}\\ B=\dfrac{3-\dfrac{1}{3^{2005}}}{2}\)
b,
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\\ 5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\\ 5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\\ 4A=5^{51}-1\\ A=\dfrac{5^{51}-1}{4}\)
c,
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2-1}\right)......\left(\dfrac{1}{100^2-1}\right)\\ A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)......\left(\dfrac{1}{10000}-1\right)\\ A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\dfrac{9999}{10000}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\cdot\dfrac{99\cdot101}{100\cdot100}\\ A=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\\ A=\dfrac{1}{100}\cdot\dfrac{101}{2}\\ A=\dfrac{101}{200}\)
17 tháng 6 2018
d,
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2^1\right)\)
Đặt \(A=B-C\)
\(\Rightarrow B=\left(2^{100}+2^{98}+...+2^2\right)vàC=\left(2^{99}+2^{97}+...+2^1\right)\)
\(B=2^{100}+2^{98}+...+2^2\\ 4B=2^{102}+2^{100}+...+2^4\\ 4B-B=\left(2^{102}+2^{100}+...+2^4\right)-\left(2^{100}+2^{98}+...+2^2\right)\\ 3B=2^{102}-2^2\\ B=\dfrac{2^{102}-2^2}{3}\left(1\right)\)
\(C=2^{99}+2^{97}+...+2^1\\ 4C=2^{101}+2^{99}+...+2^3\\ 4C-C=\left(2^{101}+2^{99}+...+2^3\right)-\left(2^{99}+2^{97}+...+2\right)\\ 3C=2^{101}-2\\ C=\dfrac{2^{101}-2}{3}\left(2\right)\)
Từ (1) và (2) ta có :
\(A=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ A=\dfrac{2^{102}-2^2-2^{101}+2}{3}\\ A=\dfrac{2^{102}-2^{101}+2}{3}\)
13 tháng 5 2022
9: \(=1-\dfrac{1}{99}+1-\dfrac{1}{100}+\dfrac{100}{101}\cdot\dfrac{1-4+3}{12}=2-\dfrac{199}{9900}=\dfrac{19601}{9900}\)
10: \(=\left(\dfrac{78}{79}+\dfrac{79}{80}+\dfrac{80}{81}\right)\cdot\dfrac{6+5+9-20}{30}=0\)
27 tháng 6 2022
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
