Giải thích các bước giải:

Bài 10: a) Ta có: A = 1992 x 1992 = (1991+1) x 1992 = 1991 x 1992 + 1992

B = 1991 x 1993 = 1991 x (1992+1) = 1991 x 1992 + 1991

Vì 1992 > 1991 ⇒ 1991 x 1992 + 1992 > 1991 x 1992 + 1991

                          hay A > B

b) Ta có: M = 1990 x 1994 = 1990 x (1992+2) = 1900 x 1992 + 1990 x 2

N = 1992 x 1992 = (1990+2) x 1992 = 1990 x 1992 + 1992 x 2

Vì 1990 < 1992 ⇒ 1990 x 2 < 1992 x 2 ⇒ 1900 x 1992 + 1990 x 2 < 1990 x 1992 + 1992 x 2

                                                              hay M < N

Bài 11: a) 52 x 64 + 520 x 7 - 52 x 34

= (52 x 64 - 52 x 34) + 520 x 7

= 52 x (64-34) + 520 x 7

= 52 x 30 + 520 x 7

= 52 x 10 x 3 + 520 x 7

= 520 x 3 + 520 x 7

= 520 x (3+7)

= 520 x 10

= 5200

b) 26 + 45 x 26 + 260 + 44 x 26

= 26 + 45 x 26 +26 x 10 + 44 x 26

= 26 x (1+45+10+44)

= 26 x 100

= 2600

c) $\frac{1995×1997-1}{1996×1995+1994}$ = $\frac{1995×(1996+1)-1}{1996×1995+1994}$ = $\frac{1995×1996+1995-1}{1996×1995+1994}$ = $\frac{1995×1996+1994}{1996×1995+1994}$ = 1

d) $\frac{2}{3×5}$ + $\frac{2}{5×7}$ + $\frac{2}{7×9}$ + $\frac{2}{9×11}$+ ... + $\frac{2}{13×15}$= $\frac{1}{3}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{9}$ + $\frac{1}{9}$ - $\frac{1}{11}$ + ... + $\frac{1}{13}$ - $\frac{1}{15}$ = $\frac{1}{3}$ - $\frac{1}{15}$ = $\frac{5}{15}$ - $\frac{1}{15}$ = $\frac{4}{15}$

e) $\frac{2}{1×2}$ + $\frac{2}{2×3}$ + $\frac{2}{3×4}$ + ... + $\frac{2}{8×9}$ + $\frac{2}{9×10}$= 2 x ($\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ + ... + $\frac{1}{8×9}$ + $\frac{1}{9×10}$) = 2 x ($\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{8}$ - $\frac{1}{9}$ + $\frac{1}{9}$ - $\frac{1}{10}$) = 2 x (1 - $\frac{1}{10}$) = 2 x $\frac{9}{10}$ = $\frac{9}{5}$

g) Đặt A = $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + ... + $\frac{1}{729}$

A = $\frac{1}{3^{1}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + ... + $\frac{1}{3^{6}}$

3 × A = 1 + $\frac{1}{3^{1}}$ + $\frac{1}{3^{2}}$ + ... + $\frac{1}{3^{5}}$

3 x A - A = (1 + $\frac{1}{3^{1}}$ + $\frac{1}{3^{2}}$ + ... + $\frac{1}{3^{5}}$) - ($\frac{1}{3^{1}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{3^{3}}$ + ... + $\frac{1}{3^{6}}$)

2 x A = 1 + $\frac{1}{3^{1}}$ + $\frac{1}{3^{2}}$ + ... + $\frac{1}{3^{5}}$ - $\frac{1}{3^{1}}$ - $\frac{1}{3^{2}}$ - $\frac{1}{3^{3}}$ - ... - $\frac{1}{3^{6}}$

2 x A = 1 - $\frac{1}{3^{6}}$

A = (1 - $\frac{1}{3^{6}}$) : 2

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