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c
\(=160:\left\{17+\left[9\cdot5-\left(14+2^3\right)\right]\right\}\)
\(=160:\left\{17+\left[45-\left(14+8\right)\right]\right\}\)
\(=160:\left\{17+\left[45-22\right]\right\}\)
\(=160:\left(17+23\right)\)
\(=160:40\)
\(=4\)
d
\(=798+100:\left[16-2\cdot\left(25-22\right)\right]\)
\(=798+100:\left(16-2\cdot3\right)\)
\(=798+100:\left(16-6\right)\)
\(=798+100:10\)
\(=798+10\)
\(=808\)
a: \(=\left(\dfrac{17}{10}+\dfrac{70}{10}-\dfrac{87}{10}\right):\left(\dfrac{23}{4}-\dfrac{11}{4}+\dfrac{9}{25}\right)\cdot\left(12,98\cdot0,25\right)+12,5\)
\(=0:\left(3+\dfrac{9}{25}\right)\cdot\left(12,98+0,25\right)+12,5\)
=12,5
b: \(=\dfrac{13}{12}\cdot\dfrac{27}{5}\cdot2\cdot\dfrac{34}{9}\cdot2\cdot\dfrac{2}{17}\)
\(=\dfrac{13}{12}\cdot2\cdot\dfrac{27}{5}\cdot\dfrac{34}{9}\cdot\dfrac{4}{17}\)
\(=\dfrac{13}{6}\cdot\dfrac{27}{5}\cdot\dfrac{8}{9}=\dfrac{8}{6}\cdot3\cdot\dfrac{13}{5}=4\cdot\dfrac{13}{5}=\dfrac{52}{5}\)
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)
a) (7x - 11)3 = 25 x 52 + 200
(7x - 11)3 = 800 + 200
(7x - 11)3 = 1000
(7x - 11)3 = 103
=> 7x - 11 = 10
=> 7x = 10 + 11
=> 7x = 21
=> x = 3
b) \(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(3\frac{1}{3}x=-13,25-16\frac{3}{4}\)
\(\frac{10}{3}x=-30\)
\(x=-9\)
\(a)\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{-11.13^7}{11^5.13^8}=\frac{-1}{11^4.13}\) (Bạn xem thử xem có sai đề không nhé)
\(c)\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3+1\right)}{2^9.3^{10}}=\frac{2.4}{3}=\frac{8}{3}\)
\(d)\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(5.4+9\right)}=\frac{8}{20+9}=\frac{8}{29}\)
\(a)\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-5}{3^2}=\frac{-5}{9}\)
\(b)\frac{\left(-11\right)\cdot13^7}{11^5\cdot13^8}=\frac{-1}{11^4\cdot13}=\frac{-1}{14641\cdot13}=\frac{-1}{190333}\)
\(c)\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}=\frac{2^{10}\left(3^{10}-3^9\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\left(3-1\right)}{2^9\cdot3^{10}}=\frac{2^{10}\cdot3^9\cdot2}{2^9\cdot3^{10}}=\frac{2\cdot2}{3}=\frac{4}{3}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97
(7.x-11)3=22*52+200
(7.x-11)3=1000 = 103
suy ra: 7x-11=10
7x=10+11
7x=21
x=21:7=3