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\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
Lời giải:
Điều kiện: $x\neq -2; x\neq -2; x\neq -8; x\neq -14$
Đề bài
$\Rightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}$
$\Rightarrow 12=x$ (thỏa mãn)
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
1: =>|1/4x^2+1/45|=1/20
=>1/4x^2+1/45=1/20 hoặc 1/4x^2+1/45=-1/20
=>1/4x^2=1/36
=>x^2=1/36:1/4=1/9
=>x=1/3 hoặc x=-1/3
2: =(x^2-3)(x^2-2x)
=x(x-2)(x^2-3)
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
a) ta có : \(\left(x-\dfrac{1}{3}\right).\left(x+\dfrac{2}{3}\right)>0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>\dfrac{1}{3}\) hoặc \(x< \dfrac{-2}{3}\)
b) \(\left(x+\dfrac{3}{5}\right).\left(x+1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-3}{5}\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-3}{5}\\x>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\-1< x< \dfrac{-3}{5}\end{matrix}\right.\) vậy \(-1< x< \dfrac{-3}{5}\)
\(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\Rightarrow x>\dfrac{1}{3}\\x+\dfrac{2}{3}>0\Rightarrow x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\Rightarrow x< \dfrac{1}{3}\\x+\dfrac{2}{3}< 0\Rightarrow x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x>-\dfrac{2}{3}\) hoặc \(x< \dfrac{1}{3}\)
\(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\Rightarrow x< -\dfrac{3}{5}\\x+1>0\Rightarrow x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\Rightarrow x>-\dfrac{3}{5}\\x+1< 0\Rightarrow x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< -\dfrac{3}{5}\)
Lời giải:
PT \(\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)
\(\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}\)
\(\Rightarrow x=12\) (thỏa mãn)
Vậy......