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4 - 1 = 3     4 - 2 = 2     3 + 1 = 4     1 + 2 = 3

3 - 1 = 2     3 - 2 = 1     4 - 3 = 1     3 - 1 = 2

2 - 1 = 1     4 - 3 = 1     4 - 1 = 3     3 - 2 = 1

dễ thế cũng hỏi

Lời giải chi tiết:

4-1  >  2

4-2  =  2

4-3  <  4-2

4-1  <  3+1

Lời giải chi tiết:

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

hơi sai đề

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Lần đầu post, mình quên mất chưa nêu câu hỏi. Nhờ các bạn chứng minh dùm 3 câu trên với, cám ơn nhiều ah!

1.\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{100}}\)

Thấy:\(\frac{1}{2^{100}}>0\Rightarrow1-\frac{1}{2^{100}}< 1\)

\(\Rightarrow A< 1\)

Ta có:\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)=A+100< 1+100=101\)

\(101>\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(1+\frac{1}{2^{100}}\right)\ge100\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)...\left(\frac{1}{2^{100}}\right)>\left(\frac{101}{100}\right)^{100}>3\)

*Cách khác:

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)\)

\(=\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Ta thấy:

\(\frac{2+1}{2}>\frac{2^2+1}{2^2}>....>\frac{2^{100}+1}{2^{100}}\)

\(\Rightarrow\frac{2+1}{2}>\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}\)

Mà \(\frac{2+1}{2}< 3\)

\(\Rightarrow\frac{2+1}{2}.\frac{2^2+1}{2^2}....\frac{2^{100}+1}{2^{100}}< 3\)

\(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)+...+\left(1+\frac{1}{2^{100}}\right)< 3\)

Bài 1:

\(\frac14+\frac23=\frac{3}{12}+\frac{8}{12}=\frac{3+8}{12}=\frac{11}{12}\)

\(\frac27+\frac23=\frac{6}{21}+\frac{14}{21}=\frac{6+14}{21}=\frac{20}{21}\)

\(\frac25+\frac13=\frac{6}{15}+\frac{5}{15}=\frac{6+5}{15}=\frac{11}{15}\)

\(\frac23+\frac12=\frac46+\frac36=\frac{4+3}{6}=\frac76\)

\(\frac13+\frac35=\frac{5}{15}+\frac{9}{15}=\frac{5+9}{15}=\frac{14}{15}\)
\(\frac45+\frac13=\frac{12}{15}+\frac{5}{15}=\frac{12+5}{15}=\frac{17}{15}\)

\(\frac18+\frac34=\frac18+\frac68=\frac{1+6}{8}=\frac78\)

\(\frac{1}{36}+\frac{5}{12}=\frac{1}{36}+\frac{15}{36}=\frac{1+15}{36}=\frac{16}{36}=\frac49\)

\(\frac13+\frac16+\frac{1}{18}=\frac{6}{18}+\frac{3}{18}+\frac{1}{18}=\frac{6+3+1}{18}=\frac{10}{18}=\frac59\)

Bài 2:

\(\frac{15}{16}-\frac{3}{16}=\frac{15-3}{16}=\frac{12}{16}=\frac34\)

\(\frac{17}{18}-\frac56=\frac{17}{18}-\frac{15}{18}=\frac{17-15}{18}=\frac{2}{18}=\frac19\)

\(\frac34-\frac49=\frac{27}{36}-\frac{16}{36}=\frac{27-16}{36}=\frac{11}{36}\)

\(\frac12-\frac25=\frac{5}{10}-\frac{4}{10}=\frac{5-4}{10}=\frac{1}{10}\)

\(\frac56-\frac{3}{10}=\frac{25}{30}-\frac{9}{30}=\frac{25-9}{30}=\frac{16}{30}=\frac{8}{15}\)

\(3-\frac13=\frac93-\frac13=\frac{9-1}{3}=\frac83\)

\(\frac45-\frac{1}{10}=\frac{8}{10}-\frac{1}{10}=\frac{7}{10}\)

\(\frac52-1=\frac{5-2}{2}=\frac32\)

\(\frac58-\frac25=\frac{25}{40}-\frac{16}{40}=\frac{25-16}{40}=\frac{9}{40}\)

không chia hết

không chia hết