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\(.1.\)
Ta có : \(f\left(x\right)=\frac{2}{3}-\frac{1}{2}\left|x-1\right|\)
Thay
+ \(f\left(0\right)=\frac{2}{3}-\frac{1}{2}\left|0-1\right|\)
\(\Rightarrow f\left(0\right)=\frac{2}{3}-\frac{1}{2}\)
\(\Rightarrow f\left(0\right)=\frac{4}{6}-\frac{3}{6}\)
\(\Rightarrow f\left(0\right)=\frac{1}{6}\)
+ \(f\left(-1\right)=\frac{2}{3}-\frac{1}{2}\left|-1-1\right|\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{1}{2}.2\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-1\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{3}{3}\)
\(\Rightarrow f\left(-1\right)=-\frac{1}{3}\)
+ \(f\left(1\right)=\frac{2}{3}-\frac{1}{2}\left|1-1\right|\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-\frac{1}{2}.0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}\)
+ \(f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}\left|\frac{3}{4}-1\right|\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}.\frac{1}{4}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{8}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{16}{24}-\frac{3}{24}=\frac{13}{24}\)
\(y=f\left(x\right)=\dfrac{5}{x-1}\)
a) ĐKXĐ khi: \(x-1\ne0\) \(\Leftrightarrow x\ne1\)
b)
\(y=f\left(-2\right)=\dfrac{5}{-2-1}=\dfrac{5}{-\left(2+1\right)}=\dfrac{5}{-3}=\dfrac{-5}{3}\)
\(y=f\left(\dfrac{1}{3}\right)=\dfrac{5}{\dfrac{1}{3}-1}=\dfrac{5}{\dfrac{1}{3}-\dfrac{3}{3}}=\dfrac{5}{\dfrac{-2}{3}}=\dfrac{-15}{2}\)
c)
* \(y=f\left(x\right)=\dfrac{5}{x-1}=-1\)
\(\Rightarrow x-1=5:\left(-1\right)\)
\(\Rightarrow x-1=-5\)
\(\Rightarrow x=-5+1\)
\(\Rightarrow x=-\left(5-1\right)\)
\(\Rightarrow x=-4\)
Vậy \(y=-1\) thì \(x=-4\)
* \(y=f\left(x\right)=\dfrac{5}{x-1}=1\)
\(\Rightarrow x-1=5:1\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=5+1\)
\(\Rightarrow x=6\)
Vậy \(y=1\) thì \(x=6\)
* \(y=f\left(x\right)=\dfrac{5}{x-1}=\dfrac{1}{5}\)
\(\Rightarrow x-1=5:\dfrac{1}{5}\)
\(\Rightarrow x-1=5.5\)
\(\Rightarrow x-1=25\)
\(\Rightarrow x=25+1\)
\(\Rightarrow x=26\)
Vậy \(y=\dfrac{1}{5}\) thì \(x=26\)
D