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a: f(0)=5
f(-1)=2+5=7
f(1)=-2+5=3
f(-2)=4+5=9
f(1/2)=-1+5=4
f(-1/2)=2+5=7
b: Khi y=5 thì -2x+5=5
=>x=0
Khi y=-3 thì -2x+5=-3
=>-2x=-8
hay x=4
Khi y=19 thì -2x+5=19
=>-2x=14
hay x=-7
\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)
Bài làm:
a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)
+ Nếu x = 6
\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)
+ Nếu x = 4
\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)
\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)
b) \(\frac{3}{2}x-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{5}{3}\)
\(\Leftrightarrow\frac{3}{2}x-\frac{1}{2}x+\frac{1}{3}=\frac{5}{3}\)
\(\Leftrightarrow x=\frac{4}{3}\)
Thay vào ta được:
\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)
\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)
\(\Leftrightarrow14y=-4\)
\(\Rightarrow y=-\frac{2}{7}\)
Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)
\(A=\dfrac{3}{2}\left|2\cdot\dfrac{1}{2}+\dfrac{1}{3}\right|-\left(-\dfrac{2}{5}\right)\cdot\dfrac{5}{2}=\dfrac{3}{2}\cdot\dfrac{4}{3}+1=2+1=3\)