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a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
=> x-3 = -1
=> x = 2
hoặc x-3 = 0
x = 3
hoặc x-3 = 1
x = 4
Vậy x\(\in\){2; 3; 4} thì (x-3)20 = (x-3)10
`@` `\text {Ans}`
`\downarrow`
`16/5 - x = 4/5 - 3/10`
`=> 16/5 - x = 1/2`
`=> x = 16/5 - 1/2`
`=> x = 27/10`
Vậy, `x = 27/10.`
\(x\times\left(x-y\right)=\frac{3}{10};y\times\left(x-y\right)=-\frac{3}{20}\)
\(\Rightarrow\frac{x}{y}=\frac{x\cdot\left(x-y\right)}{y\cdot\left(x-y\right)}=\frac{\frac{3}{10}}{-\frac{3}{20}}=\frac{3}{10}\cdot-\frac{20}{3}=-2\)
\(\Rightarrow y=-\frac{1}{2y}\Rightarrow x\cdot\left(x-y\right)=x\cdot\left(x-\right)\)
\(\left||x+3\right|-10|=20\)
\(\Rightarrow\orbr{\begin{cases}\left|x+3\right|-10=20\\\left|x+3\right|-10=-20\end{cases}}\Rightarrow\orbr{\begin{cases}\left|x+3\right|=30\\\left|x+3\right|=-10\end{cases}}\)
Ta có: \(\left|x+3\right|\ge0\forall x\)
\(\Rightarrow\left|x+3\right|\ne-10\)
\(\Rightarrow\left|x+3\right|=30\Rightarrow\orbr{\begin{cases}x+3=30\\x+3=-30\end{cases}}\Rightarrow\orbr{\begin{cases}x=27\\x=-33\end{cases}}\)
|| x + 3 | - 10 | = 20
\(\Rightarrow\orbr{\begin{cases}\left|x+3\right|-10=20\\\left|x+3\right|-10=-20\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left|x+3\right|=30\\\left|x+3\right|=-10\end{cases}}\)mà | x + 3 | ≥ 0 ∀ x
=> | x + 3 | = 30
\(\Rightarrow\orbr{\begin{cases}x+3=30\\x+3=-30\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=27\\x=-33\end{cases}}\)