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2)
\(D=\dfrac{4}{3}+\dfrac{10}{9}+\dfrac{28}{27}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3+1}{3}+\dfrac{3^2+1}{3^2}+\dfrac{3^3+1}{3^3}+...+\dfrac{3^{98}+1}{3^{98}}\\ D=\dfrac{3}{3}+\dfrac{1}{3}+\dfrac{3^2}{3^2}+\dfrac{1}{3^2}+\dfrac{3^3}{3^3}+\dfrac{1}{3^3}+...+\dfrac{3^{98}}{3^{98}}+\dfrac{1}{3^{98}}\\ D=1+\dfrac{1}{3}+1+\dfrac{1}{3^2}+1+\dfrac{1}{3^3}+...+1+\dfrac{1}{3^{98}}\\ D=\left(1+1+1+...+1\right)+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ D=98+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\)
Gọi \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\) là \(C\)
\(C=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\\ 3C=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\\ 3C-C=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{97}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{98}}\right)\\ 2C=1-\dfrac{1}{3^{98}}\\ C=\left(1-\dfrac{1}{3^{98}}\right):2\\ C=1:2-\dfrac{1}{3^{98}}:2\\ C=\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}\)
\(D=98+C=98+\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}=98\dfrac{1}{2}-\dfrac{1}{3^{98}\cdot2}< 100\)
Vậy \(D< 100\)
Ta có: \(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+...+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+...+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+...+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)
\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\)
⇒5A=15+252+...+11511⇒5A=15+252+...+11511
⇒5A−A=15+152+...+1511−11512⇒5A−A=15+152+...+1511−11512
⇒4A=15+152+...+1511−11512⇒4A=15+152+...+1511−11512
⇒20A=1+15+...+1510−11511⇒20A=1+15+...+1510−11511
⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)⇒20A−4A=(1+15+...+1510−11511)−(15+152+...+1511−11512)
⇒16A=1−12511+11512<1⇒16A=1−12511+11512<1
⇒A<116⇒A<116
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
Bài này có cần phải tính nhanh ko vậy bn?
Nếu ko thì lấy máy tính mà tính cũng đc mà
11) \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right):\dfrac{7}{5}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}\right)+\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\cdot\dfrac{5}{7}\)
= \(\dfrac{5}{7}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{4}{7}+\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{4}{7}\right)\)
= \(\dfrac{5}{7}\cdot0\)
=0
12) \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2\right)-\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\left(\dfrac{17}{3}-\dfrac{16}{9}+2-\dfrac{17}{3}+\dfrac{16}{9}\right)\)
= \(\dfrac{43}{5}\cdot2=\dfrac{43}{10}\)
11, 5/7( 1/2-1/3+1/4)+ (1/3-1/2-1/4):7/5
= 5/7.(1/2 - 1/3 + 1/4 )+( 1/3 - 1/2 - 1/4). 5/7
= 5/7.(1/2 - 2/3 + 1/4 + 1/3 - 1/2 - 1/4)
= 5/7 . -1/3
= -5/21
12, 43/5.(17/3 - 16/9 + 2)- 43/5. (17/3 - 16/9)
= 43/5.( 17/3 - 16/9 + 2 - 17/3 + 16/9)
= 43/5 . 2
= 86/5
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
Ta có :
\(A=\dfrac{1}{5^2}+\dfrac{2}{5^3}+\dfrac{3}{5^4}+.............+\dfrac{n}{5^{n+1}}+.....+\dfrac{11}{5^{12}}\)
\(\Rightarrow5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{3^3}+........+\dfrac{n}{5^n}+..........+\dfrac{11}{5^{11}}\)
\(\Rightarrow5A-A=\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+.....+\dfrac{n}{5^n}+....+\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5^2}+\dfrac{2}{5^3}+.....+\dfrac{n}{5^{n+1}}+........+\dfrac{11}{5^{12}}\right)\)\(\Rightarrow4A=\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\)
\(\Rightarrow20A=1+\dfrac{1}{5}+.........+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\)
\(\Rightarrow20A-4A=\left(1+\dfrac{1}{5}+.......+\dfrac{1}{5^{10}}-\dfrac{11}{5^{11}}\right)-\left(\dfrac{1}{5}+\dfrac{1}{5^2}+........+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}\right)\)\(\Rightarrow16A=1-\dfrac{12}{5^{11}}+\dfrac{11}{5^{12}}< 1\)
\(\Rightarrow A< \dfrac{1}{16}\rightarrowđpcm\)
1)
\(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\\ 5A=1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\\ 5A-A=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\right)\\ 4A=1-\dfrac{500}{5^{500}}\\ A=\left(1-\dfrac{500}{5^{500}}\right):4\\ A=1:4-\dfrac{500}{5^{500}}:4\\ A=\dfrac{1}{4}-\dfrac{500}{5^{500}\cdot4}< \dfrac{1}{4}< \dfrac{5}{16}\)
Vậy \(A< \dfrac{5}{16}\)