\(\dfrac{6x}{x-2}+\dfrac{12}{2-x}\)
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10 tháng 8 2023
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}=\dfrac{9}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}+\dfrac{1}{2}=2\\x=-\dfrac{3}{2}+\dfrac{1}{2}=-1\end{matrix}\right.\)
10 tháng 8 2023
\(\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{3}=\dfrac{23}{12}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{23}{12}+\dfrac{1}{3}\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
BD
20 tháng 8 2023
Sửa đề
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
10 tháng 3 2018
Nguyễn Thanh Hằng Nhã Doanh ngonhuminh nguyen thi vang mấy ban giup mk voi
21 tháng 8 2023
\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)
\(X+16=-5\)
\(X=-21\)
12 tháng 4 2022
\(A=\dfrac{x+3}{\left(x-3\right)^2}:\dfrac{12-x^2+x+x^2-9}{x\left(x-3\right)}\)
\(=\dfrac{x+3}{\left(x-3\right)^2}\cdot\dfrac{x\left(x-3\right)}{x+3}=\dfrac{x}{x-3}\)
KT
1
12 tháng 12 2021
\(=\dfrac{2}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+6-2x+4}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\dfrac{10}{\left(x-2\right)\left(x-3\right)\left(x+3\right)}\)
HT
18 tháng 8 2017
\(=\left[\dfrac{x^3}{x^2\left(x^2-4\right)}+\dfrac{12}{6\left(2-x\right)}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)
\(=\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right]:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left(\dfrac{10-x^2}{x+2}+x-2\right)\)
\(=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}:\dfrac{10-x^2+x^2+2x-2x-4}{x+2}\)
\(=-\dfrac{6}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)
\(=\dfrac{-1}{x-2}\)
= \(\dfrac{6x}{x-2}+\dfrac{-12}{x-2}\) = \(\dfrac{6x-12}{x-2}=\dfrac{6\left(x-2\right)}{x-2}=6\)
\(=6\)