[Toán đại 7] Bài tập khó

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thcsdn25

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soicon_boy_9x

Bài 1:
$\dfrac{x+5}{1995}+\dfrac{x+4}{1996}+\dfrac{x+3}{1997}=\dfrac{x+1995}{5} +\dfrac{x+1996}{4}+\dfrac{x+1997}{3}$
$\rightarrow \dfrac{x+5}{1995}+\dfrac{x+4}{1996}+\dfrac{x+3}{1997}+3=\dfrac{x+1995}{5}+\dfrac{x+1996}{4}+ \dfrac{x+1997}{3} +3=dfrac{x+1995}{5}+\dfrac{x+1996}{4}+\dfrac{x+1997}{3}+3$
$\rightarrow \dfrac{x+5}{1995}+1+\dfrac{x+4}{1996}+1+\dfrac{x+3}{1997}+1=\dfrac{x+1995}{5}+1+\dfrac{x+1996}{4}+1+\dfrac{x+1997}{3}+1$
$\rightarrow \dfrac{x+2000}{1995}+\dfrac{x+2000}{1996}+\dfrac{x+2000}{1997}=\dfrac{x+2000}{5}+\dfrac{x+2000}{4}+ \dfrac{x+2000}{3} $
$\rightarrow (x+2000)(\dfrac{1}{1995}+\dfrac{1}{1996}+\dfrac{1}{1997})=(x+2000)(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{3})$
$\rightarrow x+2000=0 \rightarrow x=-2000$
Bài 2:
Ta có $\dfrac{1}{2^2}<\dfrac{1}{1.2}$
$\dfrac{1}{3^2}<\dfrac{1}{2.3}$
...
$\dfrac{1}{100^2}<\dfrac{1}{99.100}$
$\rightarrow 1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}>1-(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}$
$\rightarrow 1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}>1-(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$\rightarrow 1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}>1-(\dfrac{1}{100})$
$\rightarrow 1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}>\dfrac{1}{100}$
 
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luffy_1998

a. $\dfrac{x+1}{x-1} \in Z$
$ \rightarrow \dfrac{x-1+2}{x-1} \in Z$
$ \rightarrow 1 + \dfrac{2}{x-1} \in Z$
$ \rightarrow x - 1 \in {\pm 1; \pm 2}$
Tự giải tiếp
b. $\dfrac{6x+5}{2x-1}$
$ \rightarrow \dfrac{6x - 3 + 8}{2x-1} \in Z$
$ \rightarrow 3 + \dfrac{8}{2x-1} \in Z$
$ \rightarrow 2x-1 \in {\pm 1; \pm 2; \pm 4; \pm 8}$
Tự giải tiếp
 
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