Câu 4 : Giải
(Có vẽ hình đó !!!)
Gọi diện tích của các hình là $\ S$.
Ta có :
$S_{MNQ}$ = 21 × 20 : 2 = 210 ( $cm^{2}$ )
$S_{MEQ}$ = 5,25 × 20 ÷ 2 = 52,5 ( $cm^{2}$ )
$S_{MNE}$ = 210 - 52,5 = 157,5 ( $cm^{2}$ )
KE = 157,5 × 2 : 21 = 15 ( $\ cm$ )
ĐS : 15 $\ cm$.
Câu 5 :
$\dfrac{1}{2}$ + $\dfrac{5}{6}$ + $\dfrac{11}{12}$ + $\dfrac{19}{20}$ + $\dfrac{29}{30}$ + $\dfrac{41}{42}$ + $\dfrac{55}{56}$ + $\dfrac{71}{72}$ + $\dfrac{89}{90}$
= ( 1 - $\dfrac{1}{2}$ ) + ( 1 - $\dfrac{1}{6}$ ) + ( 1 - $\dfrac{1}{12}$ ) + ( 1 - $\dfrac{1}{20}$ ) + ( 1 - $\dfrac{1}{30}$ ) + ( 1 - $\dfrac{1}{42}$ ) + ( 1 - $\dfrac{1}{56}$ ) + ( 1 - $\dfrac{1}{79}$ ) + ( 1 - $\dfrac{1}{90}$ )
= ( 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 ) - ( $\dfrac{1}{2}$ + $\dfrac{1}{6}$ + $\dfrac{1}{12}$ + $\dfrac{1}{20}$ + $\dfrac{1}{30}$ + $\dfrac{1}{42}$ + $\dfrac{1}{56}$ + $\dfrac{1}{79}$ + $\dfrac{1}{90}$
= ( 1 × 9 ) - ( $\dfrac{1}{1 × 2}$ + $\dfrac{1}{2 × 3}$ + $\dfrac{1}{3 × 4}$ + $\dfrac{1}{4 × 5}$ + $\dfrac{1}{5 × 6}$ + $\dfrac{1}{6 × 7}$ + $\dfrac{1}{7 × 8}$ + $\dfrac{1}{8 × 9}$ + $\dfrac{1}{9 × 10}$
= 9 - ( $\dfrac{1}{1}$ - $\dfrac{1}{2}$ + $\dfrac{1}{2}$ - $\dfrac{1}{3}$ + $\dfrac{1}{3}$ - $\dfrac{1}{4}$ + $\dfrac{1}{4}$ - $\dfrac{1}{5}$ + $\dfrac{1}{5}$ - $\dfrac{1}{6}$ + $\dfrac{1}{6}$ - $\dfrac{1}{7}$ + $\dfrac{1}{7}$ - $\dfrac{1}{8}$ + $\dfrac{1}{8}$ - $\dfrac{1}{9}$ + $\dfrac{1}{9}$ - $\dfrac{1}{10}$
= 9 - ( $\dfrac{1}{1}$ - $\dfrac{1}{10}$ )
= 9 - $\dfrac{9}{10}$
= $\dfrac{81}{10}$ = 8,1
