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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 !!! ^_^
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}
Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 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
\(\left(3^{19}.7+3^{19}.20\right).3^{10}\)
\(=\left[3^{19}.\left(7+20\right)\right].3^{10}\)
\(=\left[3^{19}.27\right].3^{10}\)
\(=\left[3^{19}.3^3\right].3^{10}\)
\(=3^{22}.3^{10}\)
\(=3^{32}\)