\(-15a+12\ge-15b+12\)
\(\Leftrightarrow-15a+12-12\ge-15b\)
\(\Leftrightarrow-15a\ge-15b\)
\(\Leftrightarrow\left(-1\right)\cdot\left(-15a\right)\le\left(-1\right)\cdot\left(-15b\right)\)
\(\Leftrightarrow15a\le15b\)
\(\Leftrightarrow a\le b\)
Vậy : \(a\le b\)
Ta có :
\(-15a+12\ge-15b+12\)\(2\)
\(\Rightarrow-15a+12+\left(-12\right)\ge-15b+12+\left(-12\right)\)
\(\Rightarrow-15a\ge-15b\)
\(\Rightarrow-15a.\frac{1}{-15}\le-15b.\frac{1}{-15}\)
\(\Rightarrow a\le b\)
\(-15a+12\ge-15b+12\)
\(\Rightarrow-15a\ge-15b\)\(\Rightarrow\left(-15a\right)-\left(-15b\right)\ge0\)
\(\Rightarrow-15a+15b\ge0\)\(\Leftrightarrow15\left(-a+b\right)\ge0\)
\(\Leftrightarrow-a+b\ge0\)\(\Leftrightarrow b-a\ge0\)\(\Leftrightarrow b\ge a\)
hay \(a\le b\)
