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Ta có: \(16\cdot A=\dfrac{16\cdot\left(4^{15}+1\right)}{4^{17}+1}\)
\(\Leftrightarrow16\cdot A=\dfrac{4^{17}+16}{4^{17}+1}=1+\dfrac{15}{4^{17}+1}\)
Ta có: \(16\cdot B=\dfrac{16\cdot\left(4^{12}+1\right)}{4^{14}+1}\)
\(\Leftrightarrow16\cdot B=\dfrac{4^{14}+16}{4^{14}+1}=1+\dfrac{15}{4^{14}+1}\)
Ta có: \(4^{17}+1>4^{14}+1\)
\(\Leftrightarrow\dfrac{15}{4^{17}+1}< \dfrac{15}{4^{14}+1}\)
\(\Leftrightarrow\dfrac{15}{4^{17}+1}+1< \dfrac{15}{4^{14}+1}+1\)
\(\Leftrightarrow16A< 16B\)
hay A<B
a) \(\dfrac{-1}{20}=\dfrac{-7}{140}\)
\(\dfrac{5}{7}=\dfrac{100}{140}\)
mà -7<100
nên \(-\dfrac{1}{20}< \dfrac{5}{7}\)
b) \(\dfrac{216}{217}< 1\)
\(1< \dfrac{1164}{1163}\)
nên \(\dfrac{216}{217}< \dfrac{1164}{1163}\)
c) \(\dfrac{-12}{17}=\dfrac{-180}{255}\)
\(\dfrac{-14}{15}=\dfrac{-238}{255}\)
mà -180>-238
nên \(-\dfrac{12}{17}>\dfrac{-14}{15}\)
d) \(\dfrac{27}{29}>0\)
\(0>-\dfrac{2727}{2929}\)
nên \(\dfrac{27}{29}>-\dfrac{2727}{2929}\)
\(a,\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b,\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14+\left(-15\right)}{21}=\dfrac{-29}{21}\\ c,\dfrac{1}{2}-\dfrac{-1}{2}-\dfrac{1}{3}=\dfrac{3+3-2}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(a.\dfrac{4}{28}+\dfrac{12}{36}=\dfrac{1}{7}+\dfrac{1}{3}=\dfrac{3}{21}+\dfrac{7}{21}=\dfrac{10}{21}\\ b.\dfrac{-12}{18}+\dfrac{-15}{21}=\dfrac{-2}{3}+\dfrac{-5}{7}=\dfrac{-14}{21}+\dfrac{-15}{21}=\dfrac{-29}{21}\\ c.\dfrac{14}{28}+\dfrac{16}{32}-\dfrac{17}{51}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{17}{51}=1-\dfrac{17}{51}=\dfrac{2}{3}\)
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
* Cách 1 :
Ta có :
\(16A=\frac{4^{17}+16}{4^{17}+1}=\frac{4^{17}+1+15}{4^{17}+1}=\frac{4^{17}+1}{4^{17}+1}+\frac{15}{4^{17}+1}=1+\frac{15}{4^{17}+1}\)
\(16B=\frac{4^{14}+16}{4^{14}+1}=\frac{4^{14}+1+15}{4^{14}+1}=\frac{4^{14}+1}{4^{14}+1}+\frac{15}{4^{14}+1}=1+\frac{15}{4^{14}+1}\)
Vì \(\frac{15}{4^{17}+1}< \frac{15}{4^{14}+1}\) nên \(1+\frac{15}{4^{17}+1}< 1+\frac{15}{4^{14}+1}\)
\(\Rightarrow\)\(16A< 16B\) hay \(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(4^2.A=\frac{4^2\left(4^{15}+1\right)}{4^{17}+1}\); \(4^2.B=\frac{4^2\left(4^{12}+1\right)}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+4^2}{4^{17}+1}\);\(4^2.B=\frac{4^{14}+4^2}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+1+4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1+4^2-1}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+1}{4^{17}+1}+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1}{4^{14}+1}+\frac{4^2-1}{4^{14}+1}\)
=> \(4^2.A=1+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=1+\frac{4^2-1}{4^{14}+1}\)
Mà \(4^{17}>4^{14}\)
=> \(4^{17}+1>4^{14}+1\)
=> \(\frac{4^2-1}{4^{17}+1}< \frac{4^2-1}{4^{14}+1}\)
=> \(1+\frac{4^2-1}{4^{17}+1}< 1+\frac{4^2-1}{4^{14}+1}\)
=> \(4^2.A< 4^2.B\)
=> \(A< B\)
Câu1:
a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)
\(=2008^2-\left(2008^2-4\right)\)
=4
b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)
c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)
=0
a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)
\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)
\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)
mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)
\(\Rightarrow A< B\)
A = \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)
3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)
3A - A = 1 - \(\dfrac{1}{3^{2023}}\)
2A = 1 - \(\dfrac{1}{3^{2023}}\) < 1
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)
B = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)
B = \(\dfrac{8}{12}\)
B = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1
2A < 2B ⇒ A < B
4 mũ 15+1/4 mũ 17 +1= 1/16+1
4 mũ 12+1/ 4 mũ 14+1= 1/16+1
suy ra 1/17=1/17
suy ra A=B
nhớ tích cho tớ nhé