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1/100-1/100.99-1/99.98-.....1/3.2-1/2.1
= 1/100-(1/100.99+1/99.98+.....+1/3.2+1/2.1)
=1/100-(1/1-1/2+1/2-1/3+...+1/98-1/99+1/99-1/100)
=1/100-(1/1-1/100)
=1/100-99/100
=-98/100
=-49/50
\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}+......+\frac{1}{2.1}\)
\(= \frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}\right)\)
\(= \frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}\right)\)
\(= \frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(= \frac{1}{99}-\frac{98}{99}\)
\(= \frac{-97}{99}\)
\(\dfrac{-1}{100\cdot99}+\dfrac{-1}{99\cdot98}+\dfrac{-1}{98\cdot97}+...+\dfrac{-1}{3\cdot2}+\dfrac{-1}{2\cdot1}\\ \left(-1\right)\cdot\left(\dfrac{1}{100\cdot99}+\dfrac{1}{99\cdot98}+\dfrac{1}{98\cdot97}+...+\dfrac{1}{3\cdot2}+\dfrac{1}{2\cdot1}\right)\\ =\left(-1\right)\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\\ =\left(-1\right)\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =\left(-1\right)\cdot\left(1-\dfrac{1}{100}\right)\\ =\left(-1\right)\cdot\dfrac{99}{100}\\ =\dfrac{-99}{100}\)
Câu 1:
[(4x+28).3+5.5]:5=35
[(4x+28).3+5.5]=35.5
(4x+28).3+25=175
(4x+28).3=175-25
(4x+28).3=150
4x+28=150:3
4x+28=50
4x=50-28
4x=22
x=22:4
x=5,5
a.\([\)(4x+28).3+5.5\(]\):5=35\(\Leftrightarrow\)4(x+7).3+25=175\(\Leftrightarrow\)4(x+7).3=150\(\Leftrightarrow\)4.(x+7)=50\(\Leftrightarrow\)x+7=\(\frac{25}{2}\)\(\Leftrightarrow\)x=\(\frac{11}{2}\)
b.720:\([\)41-(2x-5)\(]\)=40\(\Leftrightarrow\)41-(2x-5)=18\(\Leftrightarrow\)2x-5=23\(\Leftrightarrow\)x=14
c.3x+8x-30=25\(\Leftrightarrow\)11x=55\(\Leftrightarrow\)x=5
C= \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\) ( viet nguoc lai cho de nhin)
= \(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
= \(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
= \(-\frac{49}{50}\)
C = 1/100 - 1/100.99 - 1/99.98 - 1/98.97 - .... - 1/3.2 - 1/2.1
\(C=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{2.1}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}+\frac{1}{98}-\frac{1}{99}+...+1-\frac{1}{2}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{100}-\frac{1}{2}\right)=-\frac{1}{2}\)
\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x}.2+\frac{2}{5}=0,5\)
\(\Rightarrow\frac{2}{5x}+\frac{2}{x}+\frac{2}{5}=\frac{1}{2}\)
\(\Rightarrow2.\left(\frac{1}{5x}+\frac{1}{x}+\frac{1}{5}\right)=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5x}+\frac{5}{5x}+\frac{x}{5x}=\frac{1}{2}:2=\frac{1}{4}\)
\(\Rightarrow\frac{1+5+x}{5x}=\frac{1}{4}\)
\(\Rightarrow4.\left(1+5+x\right)=5x\)
\(\Rightarrow4+20+4x=5x\)
\(\Rightarrow24+4x=5x\)
\(\Rightarrow5x-4x=24\)
\(\Rightarrow x=24\)
\(A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}+\frac{1}{100}-1=\frac{1}{50}-1=-\frac{49}{50}\)
Đặt biểu thức trên là C
\(\Rightarrow\)\(C=\frac{1}{100}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(C=\frac{1}{100}-\frac{1}{99}-\frac{1}{98}-...-\frac{1}{3}-\frac{1}{2}-\frac{1}{2}-\frac{1}{1}\)
\(C=\frac{1}{100}-\frac{1}{99}-1\)