Đáp án:
Giải thích các bước giải:
Bài 1 :
a) $\frac{4}{3×7}$ + $\frac{4}{7× 11}$ + $\frac{4}{11×15}$ + $\frac{4}{15×19}$ + $\frac{4}{19×23}$ + $\frac{4}{23×27}$
= $\frac{1}{3}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{11}$ + $\frac{1}{11}$ - $\frac{1}{15}$ + $\frac{1}{15}$ - $\frac{1}{19}$ + $\frac{1}{19}$ - $\frac{1}{23}$ + $\frac{1}{23}$ - $\frac{1}{27}$
= $\frac{1}{3}$ - $\frac{1}{27}$ = $\frac{8}{27}$
b) $\frac{2}{3×5}$ + $\frac{2}{5×7}$ + $\frac{2}{7×9}$ + $\frac{2}{9×11}$ + $\frac{2}{11×13}$ + $\frac{2}{13×15}$ + $\frac{2}{1×2}$ + $\frac{2}{2×3}$ + $\frac{2}{3×4}$ + $\frac{2}{4×5}$ + $\frac{2}{5×6}$ + $\frac{2}{6×7}$ + $\frac{2}{7×8}$ + $\frac{2}{8×9}$ + $\frac{2}{9×10}$
= ($\frac{1}{3}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{7}$ + $\frac{1}{7}$ - $\frac{1}{9}$ + $\frac{1}{9}$ - $\frac{1}{11}$ + $\frac{1}{11}$ - $\frac{1}{13}$ + $\frac{1}{13}$ - $\frac{1}{15}$) + 2( $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ + $\frac{1}{4×5}$ + $\frac{1}{5×6}$ + $\frac{1}{6×7}$ + $\frac{1}{7×8}$ + $\frac{1}{8×9}$ + $\frac{1}{9×10}$ )
= ($\frac{1}{3}$ - $\frac{1}{15}$) + 2( 1-$\frac{1}{10}$)
= $\frac{2}{15}$ + $\frac{9}{5}$ = $\frac{29}{15}$
c) = 7($\frac{1}{1×5}$ + $\frac{1}{5×9}$ + $\frac{1}{9×13}$ + $\frac{1}{13×17}$ + $\frac{1}{17×21}$)
= 7($\frac{4}{1×5}$ + $\frac{4}{5×9}$ + $\frac{4}{9×13}$ + $\frac{4}{13×17}$ + $\frac{4}{17×21}$)
= 7($\frac{1}{1}$ - $\frac{1}{5}$ + $\frac{1}{5}$ - $\frac{1}{9}$ + $\frac{1}{9}$ - $\frac{1}{13}$ + $\frac{1}{13}$ - $\frac{1}{17}$ + $\frac{1}{17}$ - $\frac{1}{21}$)
= 7($\frac{1}{1}$ - $\frac{1}{21}$) = $\frac{140}{21}$ : 4 = $\frac{35}{21}$
d) = $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ +...+ $\frac{1}{10×11}$
= $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +...+ $\frac{1}{10}$ - $\frac{1}{11}$
= $\frac{1}{1}$ - $\frac{1}{10}$ = $\frac{9}{10}$
e) = (1-$\frac{1}{6}$) + (1-$\frac{1}{12}$) + ( 1- $\frac{1}{20}$) + (1-$\frac{1}{30}$ + (1- $\frac{1}{42}$) + (1- $\frac{1}{56}$) + (1- $\frac{1}{72}$) + (1-$\frac{1}{90}$)
= (1+1+1+...+1+1) - ($\frac{1}{2×3}$ + $\frac{1}{3×4}$ + $\frac{1}{4×5}$ + $\frac{1}{5×6}$ + $\frac{1}{6×7}$ + $\frac{1}{7×8}$ + $\frac{1}{8×9}$ + $\frac{1}{9×10}$)
= 8 - ( $\frac{1}{2}$ - $\frac{1}{10}$ ) = 8 + $\frac{2}{5}$ = $\frac{42}{5}$
f) S = $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + $\frac{1}{81}$ + $\frac{1}{243}$ + $\frac{1}{729}$
3S = 1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + $\frac{1}{81}$ + $\frac{1}{243}$
3S - S = (1 + $\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + $\frac{1}{81}$ + $\frac{1}{243}$) - ($\frac{1}{3}$ + $\frac{1}{9}$ + $\frac{1}{27}$ + $\frac{1}{81}$ + $\frac{1}{243}$ + $\frac{1}{729}$)
2S = $\frac{1}{1}$ - $\frac{1}{729}$ = $\frac{728}{729}$
S = $\frac{364}{729}$
Bài 3 :
S = 1+ $\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1/10}{y}$ +...+ $\frac{1}{45}$
$\frac{S}{2}$= $\frac{1}{2}$ + $\frac{1}{6}$ + $\frac{1}{12}$ +...+ $\frac{1}{90}$
$\frac{S}{2}$= $\frac{1}{1×2}$ + $\frac{1}{2×3}$ + $\frac{1}{3×4}$ +...+ $\frac{1}{9×10}$
$\frac{S}{2}$ = $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +...+ $\frac{1}{9}$ - $\frac{1}{10}$
$\frac{S}{2}$ = 1 - $\frac{1}{10}$ = $\frac{9}{10}$
S = $\frac{9}{20}$ < 2
Bài 4 :
S = $\frac{1}{4}$ + $\frac{1}{9}$ + $\frac{1}{16}$ +...+ $\frac{1}{10000}$ = $\frac{1}{2×2}$ + $\frac{1}{3×3}$ + $\frac{1}{4×4}$ +...+ $\frac{1}{100×100}$ < $\frac{1}{1×2}$ + $\frac{1}{2×3}$ $\frac{1}{3×4}$ + $\frac{1}{4×5}$ +...+ $\frac{1}{99×100}$
S < $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ +...+ $\frac{1}{99}$ - $\frac{1}{100}$
S < $\frac{1}{1}$ - $\frac{1}{100}$ = $\frac{99}{100}$ < 1
cho mik hn nha !!
