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a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
mk chỉ cần nhìn sơ qua là biết có câu dễ sao bn ko tự nghĩ đi hơi dễ rồi trừ khi bn đố tôi chục câu tiếng anh vật lí văn
\(S=\frac{\left(9\frac{3}{8}:5,2+3,4.2\frac{7}{34}\right):1\frac{9}{16}}{0,31.8\frac{2}{2}-5,61:27\frac{1}{3}}\)\(\Rightarrow S=\frac{\left(\frac{75}{8}.\frac{5}{26}+\frac{17}{5}.\frac{75}{34}\right):\frac{25}{16}}{\frac{31}{100}.9-\frac{561}{100}.\frac{3}{82}}\)\(\Rightarrow S=\frac{\left(\frac{75.5}{8.26}-\frac{17.75}{5.34}\right).\frac{16}{25}}{\frac{31.9}{100}-\frac{561.3}{100.82}}\)
\(\Rightarrow S=\frac{\left(\frac{375}{208}-\frac{15}{2}\right).\frac{16}{25}}{\frac{279}{100}-\frac{1682}{8200}}\)\(\Rightarrow S=\frac{\frac{-1185}{208}.\frac{16}{25}}{\frac{21196}{8200}}\)\(\Rightarrow S=\frac{-237}{65}:\frac{21196}{8200}\)\(\Rightarrow S=\frac{-194340}{137774}\)
\(\Rightarrow x=\frac{2}{3}S\Rightarrow x=\frac{2}{3}.\frac{-194340}{137774}\Rightarrow x=\frac{-388680}{413322}\)
\(M=\frac{23\frac{11}{15}-26\frac{13}{20}}{12^2+5^2}:\frac{1-\frac{1}{3}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2}-\frac{19}{37}\)\(\Rightarrow M=\frac{\frac{356}{15}-\frac{533}{20}}{12^2+5^2}:\frac{\frac{5}{8}}{3^2.13.2}-\frac{19}{37}\)
\(\Rightarrow M=\frac{\frac{-35}{12}}{12^2+5^2}.\frac{3^2.13.2}{\frac{5}{8}}-\frac{19}{37}\)\(\Rightarrow M=\frac{-84}{13}-\frac{19}{37}\Rightarrow M=\frac{-3355}{481}\Rightarrow15\%M=\frac{-3355}{481}.15\%\Rightarrow15\%M=\frac{-2013}{1924}\)
\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=> \(2x+\frac{6+1}{6}+\frac{12+1}{12}+....+\frac{90+1}{90}=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+10=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=0\)
=>\(2x+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}=0\)
=>\(2x+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=0\)
=> \(2x-\frac{1}{10}=0\)
=>2x=\(\frac{1}{10}\)=> x=1/20
mình có bị nhầm chỗ dấu suy ra thứ 3. đáng lẽ ra biểu thức đó cộng 8 chứ k phải cộng 10 do mình sơ ý nên bạn hãy sủa lại chỗ ấy
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\) \(\frac{89}{90}\)
\(=(1-\frac{1}{2})+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{30}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\) \(+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(=9-\frac{11}{10}\)
\(=\frac{79}{10}\)
~Học tốt nha~
Đặt : \(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}\)
\(\Leftrightarrow A=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+......+\left(1-\frac{1}{90}\right)\)
\(\Leftrightarrow A=\left(1+1+....+1\right)-\left(\frac{1}{2}+\frac{1}{6}+....+\frac{1}{90}\right)\)
\(\Leftrightarrow A=9-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(\Leftrightarrow A=9-\left(1-\frac{1}{10}\right)\)
\(\Leftrightarrow A=9-\frac{9}{10}=\frac{81}{90}\)
= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)
= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)
= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
= \(9+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)
a/\(\frac{64}{3}\)
b/\(\frac{449}{63}\)
c/\(\frac{13}{55}\)
d/\(\frac{399}{50}\)
e/\(\frac{9073}{253}\)
Học Tốt!!