`\qquad 2f(x)+3f(1-x)=4` $\ (1)$
+) Với $x=1$ ta có:
`(1)<=>2f(1)+3f(1-1)=4`
`<=>2f(1)+3f(0)=4`
`<=>3f(0)=4-2f(1)`
`<=>f(0)={4-2f(1)}/3` $\ (2)$
$\\$
+)Với $x=0$ ta có:
`(1)<=>2f(0)+3f(1-0)=4`
`<=>2f(0)+3f(1)=4`
`<=>2f(0)=4-3f(1)`
`<=>f(0)={4-3f(1)}/2` $\ (3)$
$\\$
Từ `(2);(3)=>{4-2f(1)}/3={4-3f(1)}/2`
`<=>2[4-2f(1)]=3[4-3f(1)]`
`<=>8-4f(1)=12-9f(1)`
`<=>-4f(1)+9f(1)=12-8`
`<=>5f(1)=4`
`<=>f(1)=4/ 5`
Vậy `f(1)=4/ 5`
