[Toán 7] Toán chứng minh

H

hatrucban5apt

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H

huy14112

1a.

Có :

$\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{9}<5. \dfrac{1}{5} $

$\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}< \dfrac{1}{10}.8$

Cộng 2 vế với nhau ta được

$\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{17}<1+ \dfrac{7}{10}=\dfrac{18}{10}<2$

$\longrightarrow D<2$

b.

$\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{10}>6. \dfrac{1}{10} $

$\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{17}> \dfrac{1}{17}.7$

Cộng 2 vế với nhau ta được

$\dfrac{1}{5}+\dfrac{1}{6}+....+\dfrac{1}{17}> \dfrac{6}{10}+ \dfrac{7}{17}=\dfrac{86}{85}>1$

$\longrightarrow 1<D<2$





 
P

pe_lun_hp

2^2/1x3 x 3^2/2x4 x4^2/3x5 x...x50^2/49x51

Nếu sắp xếp lại thứ tự các tích dưới mẫu em sẽ được.

$1.3.2.4.3.5...49.51 = 1.2.3^2.4^2...49^2.50.51$

Phép tính được viết lại :

$\dfrac{2^2.3^2.4^2....50^2}{1.2.3^2.4^2...49^2.50.51} = \dfrac{2^2.50^2}{1.2.50.51} = \dfrac{100}{51}$
 
0

0973573959thuy

Chúc bạn học tốt !!!

Bài 2 :

$A = \dfrac{2^2}{1.3}. \dfrac{3^2}{2.4}. \dfrac{4^2}{3.5}. ... . \dfrac{50^2}{49.51}$

$A = (\dfrac{2}{1}. \dfrac{3}{2} . \dfrac{4}{3}. ... .\dfrac{50}{49})(\dfrac{2}{3}. \dfrac{3}{4}. \dfrac{4}{5}. ... .\dfrac{50}{51})$

$A = \dfrac{50}{1}. \dfrac{2}{51} = \dfrac{100}{51}$
 
D

duc_2605

B2)Tính nhanh
$\dfrac{2^2}{1.3}. \dfrac{3^2}{2.4} . \dfrac{4^2}{3.5} .... \dfrac{50^2}{49.51}$

= $\dfrac{2.2}{1.3}. \dfrac{3.3}{2.4} . \dfrac{4.4}{3.5} .... \dfrac{50.50}{49.51}$

= $\dfrac{2.2.3.3.4.4...50.50}{1.3.2.4.3.5...48.50.49.51}$

= $\dfrac{2.2.3.3.4.4...50.50}{1.2.3.3.4.4.5.6...47.47.48.48.49.49.50.51}$

= $\dfrac{2^2.3^3.4^4....49^2.50^2}{1.2.3^2.4^2...47^2.48^2.49^2.50.51}$

= $\dfrac{2^2.50^2}{2.50.51}$

= $\dfrac{2.50.2.50}{2.50.51}$

= $\dfrac{2.50}{51}$ = [TEX]\frac{100}{51}[/TEX]
 
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