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công thức \(\frac{a}{b}< \frac{a+m}{b+m}\)
nên ta có : \(\frac{5^{12}+1}{5^{13}+1}< \frac{5^{12}+1+4}{5^{13}+1+4}\)\(=\frac{5^{12}+5}{5^{13}+5}=\frac{5.\left(5^{11}+1\right)}{5.\left(5^{12}+1\right)}=\frac{5^{11}+1}{5^{12}+1}\)
=> \(\frac{5^{12}+1}{5^{13}+1}< \frac{5^{11}+1}{5^{12}+1}\)
Giải như mà mình không chắc nha:
a) \(A=\frac{10^8+1}{10^9+1}\)và \(\frac{10^9+1}{10^{10}+1}\)
Ta có:
\(\frac{10^8+1}{10^9+1}\Leftrightarrow\frac{10^8+1}{10^8+10+1}\Leftrightarrow\frac{1}{10+1}=\frac{1}{11}\)
\(\frac{10^9+1}{10^{10}+1}=\frac{10^8+10+1}{10^8+10+10+1}=\frac{10+1}{10+10+1}=\frac{11}{21}\)
Ta có: \(\frac{1}{11}< \frac{11}{21}\) Vậy ......
b) Bạn giải tương tự nha! Lười lắm :v
Ta có: \(5^{12}< 5^{13}\)
\(\Rightarrow5^{12}-1< 5^{13}+1\)
\(\Rightarrow m=\frac{5^{12}-1}{5^{13}+1}< 1\)
\(\Rightarrow m>\frac{5^{12}-1-4}{5^{13}+1+4}\)
\(\Rightarrow m>\frac{5^{12}-5}{5^{13}+5}\)
\(\Rightarrow m>\frac{5\left(5^{11}-1\right)}{5\left(5^{12}+1\right)}\)
\(\Rightarrow m>\frac{5^{11}-1}{5^{12}+1}\)
\(\Rightarrow m>n\)
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
a) =-5/7 +7/8-2/7+1/8- -1/12+ -13/12
=(-5/7-2/7)+(7/8+1/8)-(-1/12--13/12)
=-7/7+8/8 - 12/12
= -1+1+1
=1
b)= ( -3/8+11/8)-(12/11+ -1/11)+(-3/5- 2/5)
= 1- 1 + (-1)
=-1
\(=-\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=-\frac{1.2....99}{2.3...100}.\frac{3.4....101}{2.3...100}\)
\(=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}\)
Học good
\(=-\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{100^2}\right)\)
\(=-\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}...\frac{100^2-1}{100^2}\)
\(=-\frac{1.3}{2^2}\cdot\frac{2.4}{3^2}...\frac{99.101}{100^2}\)
\(=-\frac{1.2...99}{2.3...100}\cdot\frac{3.4...101}{2.3.100}\)
\(=-\frac{1}{100}\cdot\frac{101}{2}\)
\(=-\frac{101}{200}\)
đặt A=\(\frac{5^{12}+1}{5^{13}+1}\);B=\(\frac{5^{11}+1}{5^{12}+1}\);C= \(\frac{5^{11}-1}{5^{12}-1}\)
ta có:nhân A,B,C với 5 ta đc:\(5A=\frac{5\left(5^{12}+1\right)}{5^{13}+1}=\frac{5^{13}+5}{5^{13}+1}=\frac{5^{13}+1+4}{5^{13}+1}=\frac{5^{13}+1}{5^{13}+1}+\frac{4}{5^{13}+1}=1+\frac{4}{5^{13}+1}\)
\(5B=\frac{5\left(5^{11}+1\right)}{5^{12}+1}=\frac{5^{12}+5}{5^{12}+1}=\frac{5^{12}+1+4}{5^{12}+1}=\frac{5^{12}+1}{5^{12}+1}+\frac{4}{5^{12}+1}=1+\frac{4}{5^{12}+1}\)
\(5C=\frac{5\left(5^{11}-1\right)}{5^{12}-1}=\frac{5^{12}-5}{5^{12}-1}=\frac{5^{12}-1-4}{5^{12}-1}=\frac{5^{12}-1}{5^{12}-1}-\frac{4}{5^{12}-1}=1-\frac{4}{5^{12}-1}\)
vì 513+1>512+1>512-1
=>\(\frac{4}{5^{12}-1}>\frac{4}{5^{12}+1}>\frac{4}{5^{13}+1}\)
\(\Rightarrow1+\frac{4}{5^{12}-1}>1+\frac{4}{5^{12}+1}>1+\frac{4}{5^{13}+1}\)
=>5C>5B>5A
=>C>B>A