\(.1.\)
Ta có : \(f\left(x\right)=\frac{2}{3}-\frac{1}{2}\left|x-1\right|\)
Thay
+ \(f\left(0\right)=\frac{2}{3}-\frac{1}{2}\left|0-1\right|\)
\(\Rightarrow f\left(0\right)=\frac{2}{3}-\frac{1}{2}\)
\(\Rightarrow f\left(0\right)=\frac{4}{6}-\frac{3}{6}\)
\(\Rightarrow f\left(0\right)=\frac{1}{6}\)
+ \(f\left(-1\right)=\frac{2}{3}-\frac{1}{2}\left|-1-1\right|\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{1}{2}.2\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-1\)
\(\Rightarrow f\left(-1\right)=\frac{2}{3}-\frac{3}{3}\)
\(\Rightarrow f\left(-1\right)=-\frac{1}{3}\)
+ \(f\left(1\right)=\frac{2}{3}-\frac{1}{2}\left|1-1\right|\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-\frac{1}{2}.0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}-0\)
\(\Rightarrow f\left(1\right)=\frac{2}{3}\)
+ \(f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}\left|\frac{3}{4}-1\right|\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{2}.\frac{1}{4}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{2}{3}-\frac{1}{8}\)
\(\Rightarrow f\left(\frac{3}{4}\right)=\frac{16}{24}-\frac{3}{24}=\frac{13}{24}\)
