2 phần 11: 1 phần 3 .3phần 2
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\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)
suy ra: \(3\left(x-3\right)=5\left(x+2\right)\\ < =>3x-9=5x+10\\ < =>3x-5x=10+9\\ < =>-2x=19\\ < =>x=-\dfrac{19}{2}\left(tm\right)\)
\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\)ĐKXĐ \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)
`<=> 3(x-3) =5 (x+2)`
`<=> 3x-9 = 5x+10`
`<=>3x -5x=10+9`
`<=> -2x=19`
`<=>x=-19/2`
\(\frac{1}{2}+1+\frac{3}{2}+...+\frac{n}{2}=33\)
\(\Leftrightarrow\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+...+\frac{n}{2}=33\)
\(\Leftrightarrow\frac{1+2+3+...+n}{2}=33\)
Đặt A = \(1+2+3+...+n\)
Số số hạng = \(\frac{n-1}{1}+1=n\)
Tổng = \(\frac{\left(n+1\right)\cdot n}{2}\)
=> \(\frac{\frac{\left(n+1\right)\cdot n}{2}}{2}=33\)
=> \(\frac{\left(n+1\right)\cdot n}{2}=66\)
=> \(\left(n+1\right)\cdot n=132=11\cdot12\)
=> n = 11
Vậy n = 11
\(\left(\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\left(\dfrac{9}{12}+\dfrac{8}{12}\right):\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\dfrac{17}{12}:\dfrac{17}{4}-\dfrac{3}{4}\)
\(=\dfrac{17.4}{17.12}-\dfrac{3}{4}\)
\(=\dfrac{1}{3}-\dfrac{3}{4}\)
\(=\dfrac{4}{12}-\dfrac{9}{12}\)
\(=-\dfrac{5}{12}\)
\(\frac{2}{11}:\frac{1}{3}.\frac{3}{2}\)
\(=\frac{2}{11}.3.\frac{3}{2}\)
\(=\frac{2.3.3}{11.2}\)
\(=\frac{9}{11}\)
Dấu . là dấu nhân nha!
Mk trả lời lúc 14:12
\(\frac{2}{11}:\frac{1}{3}.\frac{3}{2}\)
\(=\frac{6}{11}.\frac{3}{2}\)
\(=\frac{9}{11}\)