\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{32}>\frac{1}{15}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}\)

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{32}>\frac{1\cdot30}{15}\)

\(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{32}>2\)

@Thùy Nguyễn Thị Bích

Ta có 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64

= ( 1/2 - 1/4 ) + ( 1/8 - 1/16 ) + ( 1/32 -  1/64)

= 1/4 + 1/16 + 1/64 

= 16 + 4 + 1 /64

= 21/64 < 21/63 = 1/3 

Vậy 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3 ( đpcm ) Chúc bn hok tốt  . k mik nha 

Ta có :

\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)

\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}< \frac{1}{3}\)

\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}< \frac{1}{3}\)

\(\frac{16+4+1}{64}< \frac{1}{3}\)

\(\frac{21}{64}< \frac{1}{3}\)

=> 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3

Bài 1 :

\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)

\(\Leftrightarrow-2x-2-3+3x=4\)

\(\Leftrightarrow x=4+2+3=9\)

Bài 2 :

Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)

\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)⁽¹⁾

Lại có :

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)⁽²⁾

Từ ⁽¹⁾ và ⁽²⁾ , ta có :

\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)

Nguyen Ribi Nkok Ngok Khôi Bùi nguyễn ngọc dinh Phùng Tuệ Minh Akai Haruma buithianhtho ?Amanda? Nguyễn Thành Trương Nguyễn Ngô Minh Trí

Bài 1 : x=9

Ta có: \(\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}\)

\(\frac{1}{8}-\frac{1}{16}=\frac{2}{16}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16}\)

\(\frac{1}{32}-\frac{1}{64}=\frac{2}{64}-\frac{1}{64}=\frac{2-1}{64}=\frac{1}{64}\)

=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)

=\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}=\frac{21}{64}\)

Ta có: \(\frac{21}{64}< \frac{21}{63}=\frac{1}{3}\)

=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)

Đặt \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)

\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)

\(A+2A=1-\frac{1}{64}\)

\(3A=1-\frac{1}{64}< 1\)

=>A<1/3

=>đpcm

1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 = 21/64

Nên 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3

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