Đáp án:
Ta có: `\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}`
`<=> \sqrt{a-b+c}+\sqrt{b}=\sqrt{a}+\sqrt{c}`
`<=> (\sqrt{a-b+c}+\sqrt{b})^2=(\sqrt{a}+\sqrt{c})^2`
`<=> a-b+c+b+2\sqrt{(a-b+c)b}=a+c+2\sqrt{ac}`
`<=> \sqrt{(a-b+c)b}=\sqrt{ac}`
`<=> (a-b+c)b=ac`
`<=> (a-b)b+bc-ac=0`
`<=> (a-b)b-c(a-b)=0`
`<=> (a-b)(b-c)=0`
`<=>`$\left[ \begin{array}{l}a=b\\b=c\end{array} \right.$
`@`Xét `a=b`:
Ta có: `1/a+1/b+1/c=1`
`<=> 1/a+1/a+1/c=1`
`<=> 2c+a=ac`
`<=> a(c-1)-2(c-1)=2`
`<=> (c-1)(a-2)=2`
Vì `a,c` nguyên dương nên `a-2\ge-1; c-1\ge0`
Nên có: $\left[ \begin{array}{l}a-2=2, c-1=1\\a-2=1. c-1=2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}a=4, c=2\\a=3,c=3\end{array} \right.$
`(a=4,b=4,c=2);(a=3,b=3,c=3)`(tm $\begin{cases}\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{cases}$)
`@`Xét `b=c`:
Tương tự ta sẽ có: `(a=2,b=4,c=4);(a=3,b=3,c=3)`(tm $\begin{cases}\sqrt{a-b+c}=\sqrt{a}-\sqrt{b}+\sqrt{c}\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\end{cases}$)
Vậy các số nguyên dương `a,b,c` cần tìm là:
`(a=4,b=4,c=2);(a=3,b=3,c=3);(a=2,b=4,c=4)`
