Tìm x: 2/(x+2)(x+4)+4/(x+4)(x+8)+6/(x+8)(x+14)=x/(x+2)(x+14)
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NT
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NT
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NN
14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
31 tháng 5 2022
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+14\right)}=\dfrac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
=>x=12
IM
14 tháng 9 2016
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow x+16-x-2=x\)
\(\Rightarrow x=14\)
16 tháng 2 2022
\(8x-48+4x-12-14=-x+4\)
\(\Leftrightarrow12x-75=-x+4\Leftrightarrow13x=79\Leftrightarrow x=\dfrac{79}{13}\)
\(-7\left(8-x\right)-6\left(x+9\right)=20-x\Leftrightarrow-56+7x-6x-54=20-x\)
\(\Leftrightarrow2x=130\Leftrightarrow x=65\)
\(9x-63-80+60x=-7x+15\Leftrightarrow76x=158\Leftrightarrow x=\dfrac{79}{38}\)
\(-96-16x-60+30x=-40x-16\Leftrightarrow54x=140\Leftrightarrow x=\dfrac{70}{27}\)
\(17x-102-14x-28=4x-24-2x+4\Leftrightarrow x=110\)
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
<=>x = 12
\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=12\)
Vậy \(x=12\)