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2a) \(\frac{1979.1979+1980.21+1958}{1980.1979-1979.1978}=\frac{1979.1979+1979.21+21+1958}{1979\left(1980-1978\right)}=\frac{1979\left(1979+21+1\right)}{1979.2}=\frac{1979.2001}{1979.2}=\frac{2001}{2}\)
b) \(\frac{5^2.6^{11}.16^2+6^2.12^6.15^2}{2.6^{12}.10^4-81^2.960^3}=\frac{5^2.6^{11}.\left(2^4\right)^2+6^2.12^6.15^2}{2.6^{12}.2^4.5^4-9^4.960^3}=\frac{5^2.6^{11}.2^8+2^2.3^2.3^6.4^6.3^2.5^2}{2^5.2^{12}.3^{12}.5^4-9^4.960^3}=\frac{5^2.2^{11}.3^{11}.2^8+2^2.3^{10}.2^{12}.5^2}{2^{17}.3^{12}.5^4-3^8.2^6.3.5}=\frac{5^2.2^{19}.3^{11}+2^{14}.3^{10}.5^2}{2^{17}.3^{12}.5^4-3^9.2^6.5}\)
\(\frac{5^2\left(2^{19}.3^{11}+2^{14}.3^{10}\right)}{3^9\left(2^{17}.3^3.5^4-1.2^5.5\right)}=\frac{5^2\left(2^{14}\left(2^5+1\right)+3^{10}\left(3+1\right)\right)}{3^9.\left(2211839840\right)}=\frac{5^2\left(2^{14}.33+3^{10}.4\right)}{3^9.2211839840}=\frac{19421700}{3^9.2211839840}\)
3) a) 6n+99 chia hết cho 3n+4
=> 2(3n+4)+91 chia hết cho 3n+4
=> 91 chia hết cho 3n+4
=> 3n+4 thuộc Ư(91)=1;7;13;91
=> n thuộc 1;3;29
Trả lời:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2006}\)
\(\Rightarrow5x+6=2006\)
\(\Rightarrow5x=2000\)
\(\Rightarrow x=400\)
Vậy x = 400
Trả lời:
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)\(\frac{5}{8}\)
Đặt \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\), ta được : \(\frac{x}{2008}-A=\frac{5}{8}\) (*)
\(\Rightarrow A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(\Rightarrow A=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Thay A vào (*) , ta có:
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
Vậy x = 2008
a: =>5x-30=25
hay x=11