NHỜ THẦY CÔ GIẢI GIÚP EM BÀI TOÁN SAU. E CMAR ƠN NHIỀUa/ Không tính giá trijhayx biến đổi và so sánh 2 biểu thức sau:
M= 1/33+ 1/34+1/35+1/36 và N = 1/10.Gợi ý có: 1/33+ 1/34+1/35+1/36 >? 1/36+1/36+1/36+1/36 = 1/9 > 1/10
b/hãy so sánh A và B biết : A = 1+ 1/2+1/3+1/4+…+1/15+1/16 và B = 3
$a$)
Ta có: $\dfrac{1}{33} > \dfrac{1}{36}$
$\dfrac{1}{34} > \dfrac{1}{36}$
$\dfrac{1}{35} > \dfrac{1}{36}$
$⇒ M = \dfrac{1}{33} + \dfrac{1}{34} + \dfrac{1}{35} + \dfrac{1}{36} > \dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} = \dfrac{1}{36} \times 4 = \dfrac{1}{9}$
Mà $\dfrac{1}{9} > \dfrac{1}{10}$
$⇒ M > \dfrac{1}{10}$
$b$) `A=1+1/2+1/3+1/4+1/5+….+1/{15} + 1/{16}`
`A = 1 + (1/2 + 1/3 + 1/4) + (1/5+1/6+1/7+1/8) + (1/9+1/{10}+1/{11}+1/{12})+ (1/{13} + 1/{14} + 1/{15} + 1/{16})`
`A > 1 + (1/2 + 1/3 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + (1/{12} + 1/{12} + 1/{12} + 1/{12}) + (1/{16} + 1/{16} + 1/{16} + 1/{16})`
` A > 1 + (1/2 + 1/3 + 1/4) + 1/8 \times 4 + 1/{12} \times 4 + \frac{1}{16} \times 4`
`A > 1 + 2 \times (1/2+1/3+1/4)`
`A > 1 + 2 \times \frac{13}{12}`
`A > 1 + \frac{13}{6}`
`A > \frac{19}{6} > \frac{18}{6} =3`
$\text{1) M=$\dfrac{1}{33}$+….+$\dfrac{1}{36}$}$
$\text{Xét M:$\dfrac{1}{33}$> $\dfrac{1}{36}$}$
$\text{…..(tất cả đều lớn hơn $\dfrac{1}{36}$)}$
$\text{$\dfrac{1}{36}$= $\dfrac{1}{36}$}$
$\text{⇒M>4×$\dfrac{1}{36}$=$\dfrac{1}{9}$}$
$⇒M>N$
$\text{2) A=1+$\dfrac{1}{2}$+…+ $\dfrac{1}{16}$}$
$\text{⇒A=1+$\dfrac{1}{2}$+( $\dfrac{1}{3}$+$\dfrac{1}{4}$)+( $\dfrac{1}{5}$+..+ $\dfrac{1}{8}$)+($\dfrac{1}{9}$+..+$\dfrac{1}{12}$)+( $\dfrac{1}{13}$+..+ $\dfrac{1}{16}$)}$
$\text{⇒A=1+ $\dfrac{1}{2}$+S1+S2+S3+S4}$
$\text{Xét S1: $\dfrac{1}{3}$>$\frac{1}{4}$}$
$\text{$\dfrac{1}{4}$=$\frac{1}{4}$}$
$\text{⇒S1>2×$\dfrac{1}{4}$=$\dfrac{1}{2}$}$
$\text{Xét S2:$\dfrac{1}{5}$> $\dfrac{1}{8}$}$
$\text{……(tất cả đều lớn hơn $\dfrac{1}{8}$}$
$\text{$\dfrac{1}{8}$= $\dfrac{1}{8}$}$
$\text{⇒S2>4×$\dfrac{1}{8}$=$\dfrac{1}{2}$}$
$\text{Xét S3:$\dfrac{1}{9}$>$\dfrac{1}{12}$}$
$\text{……(tất cả đều lớn hơn $\dfrac{1}{12}$}$
$\text{$\dfrac{1}{12}$=$\dfrac{1}{12}$}$
$\text{⇒S3> $\dfrac{1}{12}$×4= $\dfrac{1}{3}$}$
$\text{Xét S4:$\dfrac{1}{13}$> $\dfrac{1}{16}$}$
$\text{……(tất cả đều lớn hơn $\dfrac{1}{16}$)}$
$\text{$\dfrac{1}{16}$=$\dfrac{1}{16}$}$
$\text{⇒S4> $\dfrac{1}{16}$ ×4=$\dfrac{1}{4}$}$
$\text{Do đó A=1+$\dfrac{1}{2}$+S1+S2+S3+S4>1+$\dfrac{1}{2}$+$\dfrac{1}{2}$+ $\dfrac{1}{2}$+$\dfrac{1}{3}$+$\dfrac{1}{4}$}$
$\text{⇒A> $\dfrac{37}{12}$> $\dfrac{36}{12}$=3}$
$⇒A>B$