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\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
\(7.3^x+20.3^x=3^{25}\)
\(3^x.\left(7+20\right)=3^{25}\)
\(3^x.3^3=3^{25}\)
\(3^x=3^{22}\)
\(\Rightarrow x=22\)
7.3x+20.3x=325
\(\Leftrightarrow\)3x . ( 7 + 20 ) = 325
\(\Leftrightarrow\)3x . 27 = 325
\(\Leftrightarrow\)3x . 33 = 325
\(\Leftrightarrow\)3x = 325 : 33
\(\Leftrightarrow\)3x = 325 - 3
\(\Leftrightarrow\)3x = 323
\(\Leftrightarrow\)x = 23
Vậy x = 23
C = \(\frac{25^{28}+25^{24}+...+25^4+25^0}{25^{30}+25^{28}+...+25^2+25^0}\)
= \(\frac{25^{28}+25^{24}+...+25^2+25^0}{\left(25^{30}+25^{26}+...+25^2\right)+\left(25^{28}+25^{24}+...+25^0\right)}\)
= \(\frac{25^{28}+25^{24}+...+25^0}{25^2\left(25^{28}+25^{24}+...+1\right)+\left(25^{28}+25^{24}+...+1\right)}\)
= \(\frac{25^{28}+25^{24}+...+1}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}=\frac{1}{626}\)
Cho 10 ****
Đặt \(A=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\cdots+\frac{25}{25\cdot31}\)
\(A=5\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{26\cdot31}\right)\)
\(A=5\cdot\left(\frac11-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\cdot\left(\frac11-\frac{1}{31}\right)\)
\(A=5\cdot\frac{30}{31}\)
\(A=\frac{150}{31}\)
Vậy \(A=\frac{150}{31}\)
\(\frac{25}{1*6}+\frac{25}{6*11}+...+\frac{25}{26*31}\)
=\(\frac{25}{1}-\frac{25}{6}+\frac{25}{6}-\frac{25}{11}+...+\frac{25}{26}-\frac{25}{31}\)
=\(\frac{25}{1}-\frac{25}{31}\)
=\(\frac{775}{31}-\frac{25}{31}\)
=\(\frac{750}{31}\)