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a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
a) \(\frac{x}{5}=\frac{2}{5}\Rightarrow x=\frac{2\cdot5}{5}=2\)
b)\(\frac{3}{8}=\frac{6}{x}\Rightarrow x=\frac{6\cdot8}{3}=16\)
c)\(\frac{1}{9}=\frac{x}{27}\Rightarrow x=\frac{27}{9}=3\)
d) \(\frac{4}{x}=\frac{8}{6}\Rightarrow x=\frac{4\cdot6}{8}=3\)
e)
\(\frac{3}{x-5}=-\frac{4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow3x+6+4x-20=0\\ \Leftrightarrow7x-14=0\\ \Leftrightarrow x=2\)
\(g,\frac{x}{-2}=-\frac{8}{x}\Rightarrow x^2=16\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy có 2 giá trị của x là 4 ; -4
a) Ta có: \(\frac{x}{5}=\frac{2}{5}\)
\(\Leftrightarrow x=\frac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\frac{3}{8}=\frac{6}{x}\)
\(\Leftrightarrow x=\frac{6\cdot8}{3}=\frac{48}{3}=16\)
Vậy: x=16
c) Ta có: \(\frac{1}{9}=\frac{x}{27}\)
\(\Leftrightarrow x=\frac{1\cdot27}{9}=\frac{27}{9}=3\)
Vậy: x=3
d) Ta có: \(\frac{4}{x}=\frac{8}{6}\)
\(\Leftrightarrow x=\frac{4\cdot6}{8}=\frac{24}{8}=3\)
Vậy: x=3
e) Ta có: \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Leftrightarrow3\cdot\left(x+2\right)=-4\cdot\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow3x+6+4x-20=0\)
\(\Leftrightarrow7x-14=0\)
\(\Leftrightarrow7x=14\)
hay x=2
Vậy: x=2
g) Sửa đề: \(\frac{x}{-2}=\frac{-8}{x}\)
Ta có: \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Leftrightarrow x^2=\left(-8\right)\cdot\left(-2\right)=16\)
hay x∈{4;-4}
Vậy: x∈{4;-4}
a) 2 - ( \(5\frac{3}{8}\)x X - \(\frac{5}{24}\)) = \(\frac{5}{12}\)
\(5\frac{3}{8}\)x X - \(\frac{5}{24}\)= \(\frac{19}{12}\)
\(5\frac{3}{8}\)x X = \(\frac{43}{24}\)
X = \(\frac{1}{3}\)
b) \(1\frac{2}{9}\): ( \(3\frac{1}{3}\)x X + \(\frac{1}{6}\)) = \(\frac{22}{23}\)
\(3\frac{1}{3}\)x X + \(\frac{1}{6}\) = \(\frac{23}{18}\)
\(3\frac{1}{3}\)x X = \(\frac{10}{9}\)
X =\(\frac{1}{3}\)
C) \(\frac{4}{5}\)x X - \(\frac{1}{2}\)x X + \(\frac{3}{4}\)x X = \(\frac{7}{40}\)
( \(\frac{4}{5}-\frac{1}{2}+\frac{3}{4}\)) x X = \(\frac{7}{40}\)
\(\frac{21}{20}\) x X = \(\frac{7}{40}\)
X =\(\frac{1}{6}\)
Câu b
\(5\cdot3^x=8\cdot3^9+7\cdot27^3\)
\(\Leftrightarrow5\cdot3^x=8\cdot3^9+7\cdot\left(3^3\right)^3\)
\(\Leftrightarrow5\cdot3^x=8\cdot3^9+7\cdot^9\)
\(\Leftrightarrow5\cdot3^x=15\cdot3^9\)
\(\Leftrightarrow3^x=3\cdot3^9=3^{10}\)
Vậy \(x=10\)
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