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\(x-40\%x=3,6\)
\(\Rightarrow100\%x-40\%x=3,6\)
\(\Rightarrow60\%x=3,6\)
\(\Rightarrow\frac{60}{100}x=3,6\)
\(\Rightarrow x=6\)
\(3\frac{2}{7}x-\frac{1}{3}=-2\frac{3}{4}\)
\(\Rightarrow\frac{23}{7}x-\frac{1}{3}=-\frac{11}{4}\)
\(\Rightarrow\frac{23}{7}x=-\frac{33}{12}+\frac{4}{12}\)
\(\Rightarrow\frac{23}{7}x=\frac{29}{12}\)
\(\Rightarrow x=\frac{29}{12}:\frac{23}{7}=\frac{203}{276}\)
\(\frac{2}{x}+\frac{1}{12}=\frac{3}{10}\)
\(\frac{2}{x}=\frac{3}{10}-\frac{1}{12}=\frac{13}{60}\)
\(13x=2\cdot60\)
\(13x=120\)
\(x=\frac{120}{13}\)
\(\frac{2}{x}+\frac{1}{12}=\frac{3}{10}\)
\(\Rightarrow\frac{2}{x}=\frac{3}{10}-\frac{1}{12}=\frac{13}{60}\)
\(\Rightarrow120=13x\)
\(\Rightarrow x=\frac{120}{13}\)
a.2/3 + 1/3 - x= 3/5
=> 1-x = 3/5
=> x = 1-3/5= 2/5
b. 1/8/15 - 2/3x = 0,2
=> 2/3x= 23/15 - 1/5= 4/3
=> x= 4/3 : 2/3=2
c. 2/3x -3/2x = 5/12
=> x( 2/3 - 3/2) = 5/12
=> x. -5/6 = 5/12
=> x= 5/12 : -5/6
=> x= -1/2
a) 2/3 + 1/3 - x = 3/5
=> 1 - x = 3/5
=> x = 1 - 3/5
x = 2/5
b) \(1\frac{8}{15}-\frac{2}{3}x=0,2\)
=> 23/15 - 2/3x = 0,2
=> 2/3x = 23/15 - 0,2
2/3x = 1,333333333
=>x = 1,333333333 : 2/3
x = 2
c) ???
\(\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+....+100}\)
Các bạn làm ơn giải giúp mình với
\(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)
\(=\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+...+\frac{1}{\frac{100.101}{2}}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{100.101}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=2.\frac{99}{202}\)
\(=\frac{99}{101}\)
Trên máy mk hiển thị , câu hỏi này 4 phút nữa mới chính thức xuất hiện ,,, máy bị j hay do câu hỏi ak ??
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
khoan đã bạn chép nhầm đề rồi thì phải số 1 kia không có dấu gì à?
Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
\(\Rightarrow A=\frac{98}{303}\div2=\frac{49}{303}\)
\(G=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..............+\frac{1}{3^{100}}\)
\(3G=1+\frac{1}{3}+\frac{1}{3^2}+...............+\frac{1}{3^{99}}\)
\(3G-G=\left(1+\frac{1}{3}+\frac{1}{3^2}+..........+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...............+\frac{1}{3^{100}}\right)\)
\(2G=1-\frac{1}{3^{100}}\)
\(\Rightarrow G=\left(1-\frac{1}{3^{100}}\right):2\)