(3^19.7+3^19.20).3^10
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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..........+\frac{1}{19.20}-\frac{x}{40}=\frac{3}{-10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow1-\frac{1}{20}-\frac{x}{40}=\frac{-3}{10}\)
\(\Rightarrow\frac{40}{40}-\frac{2}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{38}{40}-\frac{x}{40}=\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}-\frac{-12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{38}{40}+\frac{12}{40}\)
\(\Rightarrow\frac{x}{40}=\frac{50}{40}\)
\(\Rightarrow x=50\)
Vậy x = 50
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{19\cdot20}-\frac{x}{40}=\frac{-3}{10}\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{19}-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(1-\frac{1}{20}-\frac{x}{40}=\frac{3}{-10}\)
\(\frac{x}{40}=1-\frac{1}{20}-\frac{3}{-10}=1\frac{1}{4}=\frac{5}{4}\)
\(\frac{x}{40}=\frac{5}{4}\Rightarrow x=\frac{40\cdot5}{4}=50\)
\(\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{19.20}\)
\(=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=3.\left(1-\frac{1}{20}\right)\)
\(=3.\frac{19}{20}=\frac{57}{20}\)
Ủng hộ mk nha !!! ^_^
A x 3 = 1.2.3 + 2.3.3 + ...+ 19.20.3
A x 3 = 1.2.( 3 - 1) + 2.3.(4 - 1) + ...+ 19.20.( 21-18)
A x 3 = ( 1.2.3 + 2.3.4 + ....+ 19.20.21) - ( 0.1.2 + 1.2.3 + ....+ 18.19.20)
=> A x 3 = 19 x 20 x 21 = 7890
\(3A=1.2.3+2.3.3+....+19.20.3\)
\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+....+19.20\left(21-18\right)\)
\(3A=1.2.3-1.2.0+2.3.4-2.3.1+....+19.20.21-19.20.18\)
ta có thể thấy 1.2.3 = 2.3.1 và những con số khác cũng vậy
\(\Rightarrow3A=19.20.21\)
mink nghĩ vậy bạn ạ, sai đừng trách mink nha
ta có :
A.3=1.2.3+2.3.3+3.4.3+.....+19.20.3
A*3=1.2.3+2.3.(4-1)+3.4.(5-2)+...+19.20. (21-18)
A*3=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
A.3=19.20.21
A.3=7980
Mình làm đúng rồi đó k đúng cho mình nha
Tìm x
a,\(\frac{5}{4}x=\frac{2}{3}-1,5=\frac{2}{3}-\frac{3}{2}=\frac{4}{6}-\frac{9}{6}=\frac{-5}{6}\)
\(x=\frac{-5}{6}:\frac{5}{4}=\frac{-5}{6}.\frac{4}{5}=\frac{5.\left(-1\right).2.2}{2.3.5}=\frac{-2}{3}\)
b, \(\frac{2}{7}x-\frac{3}{2}x=\frac{-21}{4}.\frac{2}{7}=\frac{-3}{2}\)
\(x.\left(\frac{2}{7}-\frac{3}{2}\right)=\frac{-3}{2}\)
\(x.\frac{-17}{14}=\frac{-3}{2}\)
\(x=\frac{-3}{2}:\frac{-17}{14}=\frac{-3}{2}.\frac{14}{-17}=\frac{21}{17}\)
câu a phải là như z ms làm được bn ơi
A = 31.3+33.5+...+319.2031.3+13.5+...+319.20\frac{3}{1.3}+\frac{1}{3.5}+...+\frac{3}{19.20}
\frac{3}{1.2.3}+\frac{3}{2.3.4}+...+\frac{3}{49.50.51}
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
\(\frac{\left(3^{19}.7+3^{19}.20\right)}{3^{10}}=\frac{3^{19}.3^3}{3^{10}}=\frac{3^{22}}{3^{10}}=3^{12}\)
P/S: nhấn vào câu hỏi tương tự cx đc đó bn
\(\left(3^{19}.7+3^{19}.20\right).3^{10}\)
\(=\left[3^{19}.\left(7+20\right)\right].3^{10}\)
\(=\left[3^{19}.27\right].3^{10}\)
\(=\left[3^{19}.3^3\right].3^{10}\)
\(=3^{22}.3^{10}\)
\(=3^{32}\)