a; 7/2+33/7-75/14=20/7

b: 9/2+1/2:11/2=9/2+1/11=101/22

Đăt A = \(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+......+\frac{1}{7^{100}}\)

\(\Rightarrow7A=1+\frac{1}{7}+\frac{1}{7^2}+.....+\frac{1}{7^{100}}\)

\(\Rightarrow7A-A=1-\frac{1}{7^{100}}\)

\(\Rightarrow6A=1-\frac{1}{7^{100}}\)

\(\Rightarrow A=\frac{1-\frac{1}{7^{100}}}{6}\)

a) Cách 1:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = 8 + \frac{7}{3} - \frac{3}{5} - 5 - \frac{2}{5} - \frac{{10}}{3} + 2\\ = (8 - 5 + 2) + (\frac{7}{3} - \frac{{10}}{3}) - (\frac{3}{5} + \frac{2}{5})\\ = 5 + \frac{{ - 3}}{3} - \frac{5}{5}\\ = 5 + ( - 1) - 1\\ = 3\end{array}\)

Cách 2:

\(\begin{array}{l}(8 + 2\frac{1}{3} - \frac{3}{5}) - (5 + 0,4) - (3\frac{1}{3} - 2)\\ = (8 + \frac{7}{3} - \frac{3}{5}) - (5 + \frac{4}{{10}}) - (\frac{{10}}{3} - 2)\\ = (\frac{{120}}{{15}} + \frac{{35}}{{15}} - \frac{9}{{15}}) - (\frac{{25}}{5} + \frac{2}{5}) - (\frac{{10}}{3} - \frac{6}{3})\\ = \frac{{146}}{{15}} - \frac{{27}}{5} - \frac{4}{3}\\ = \frac{{146}}{{15}} - \frac{{81}}{{15}} - \frac{{20}}{{15}}\\ = \frac{{45}}{{15}}\\ = 3\end{array}\)

b)

\(\begin{array}{l}(7 - \frac{1}{2} - \frac{3}{4}):(5 - \frac{1}{4} - \frac{5}{8})\\ = (\frac{{28}}{4} - \frac{2}{4} - \frac{3}{4}):(\frac{{40}}{8} - \frac{2}{8} - \frac{5}{8})\\ = \frac{{23}}{4}:\frac{{33}}{8}\\ = \frac{{23}}{4}.\frac{8}{{33}}\\ = \frac{{46}}{{33}}\end{array}\)

Xét bài toán phụ sau:

Nếu \(a+b+c=0\Leftrightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)  \(\left(a,b,c\ne0\right)\)

Thật vậy

Ta có: \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{a+b+c}{abc}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\cdot\frac{0}{abc}}\)

\(=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right|\)

Bài toán được chứng minh

Quay trở lại, ta sẽ áp dụng bài toán phụ vào bài chính:

Ta có: \(P=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}+...+\sqrt{\frac{1}{2^2}+\frac{1}{779^2}+\frac{1}{801^2}}\)

Vì \(2+1+\left(-3\right)=0\) nên:

\(\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{3^2}}=\sqrt{\frac{1}{2^2}+\frac{1}{1^2}+\frac{1}{\left(-3\right)^2}}=\sqrt{\left(\frac{1}{2}+\frac{1}{1}-\frac{1}{3}\right)^2}=\frac{1}{2}+1-\frac{1}{3}\)

Tương tự ta tính được:

\(\sqrt{\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{5^2}}=\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\) ; ... ; \(\sqrt{\frac{1}{2^2}+\frac{1}{799^2}+\frac{1}{801^2}}=\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(\Rightarrow P=\frac{1}{2}+1-\frac{1}{3}+\frac{1}{2}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2}+\frac{1}{799}-\frac{1}{801}\)

\(=\frac{1}{2}\cdot400+\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{799}-\frac{1}{801}\right)\)

\(=200+\frac{800}{801}=\frac{161000}{801}=\frac{a}{b}\Rightarrow\hept{\begin{cases}a=161000\\b=801\end{cases}}\)

\(\Rightarrow Q=161000-801\cdot200=800\)

\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)

\(\Leftrightarrow-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}\)

\(\Leftrightarrow-\frac{17}{21}.\frac{20}{17}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x< \frac{1}{84}\)

\(\Leftrightarrow-\frac{80}{84}< x< \frac{1}{84}\)

\(\Leftrightarrow-80< x< 1\Leftrightarrow x\in\left\{-79;-78;...;0\right\}\)

mà để Giá trị nguyên lớn nhất của x

\(\Rightarrow x=-1\)

A = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ... + 1/100

Ta đổi A = 2-1+1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/99-1/100

A= 2 - 1 - 1/100 =200/100 -100/100 - 1/100

A= 99/100

Cảm ơn bạn Kudo Shinichi, nhưng 

1=2-1 ->ok

1/2=1-1/2 ->ok

1/3=1/2-1/3 -> sai 

vì 1/2-1/3=1/6

a) \(\frac{1}{9}+3,25+5\frac{3}{16}+4\frac{1}{3}+2,8+0,5=\frac{1}{9}+\frac{13}{4}+\frac{83}{16}+\frac{13}{3}+\frac{14}{5}+\frac{1}{2}\)

\(=\frac{11651}{720}\)

B) \(2\frac{1}{3}+0,45+4,25+\frac{1}{81}+6\frac{8}{27}=\frac{7}{3}+\frac{9}{20}+\frac{17}{4}+\frac{1}{81}+\frac{170}{27}\)

\(=\frac{10807}{810}\)

C) \(1,25+2\frac{1}{4}+4\frac{2}{5}+0,3+2,14+4\frac{1}{8}=\frac{5}{4}+\frac{9}{4}+\frac{22}{5}+\frac{3}{10}+\frac{107}{50}+\frac{33}{8}\)

\(=\frac{2893}{200}\)

CHÚC BN HỌC TỐT!!!!!