cho $\frac{a+3b-c}{c}$= $\frac{-a+b+3c}{a}$= $\frac{a-b+3c}{b}$ tính P= (3 +$\frac{a}{b}$).(3+ $\frac{b}{c}$).(3+ $\frac{c}{a}$ )
cho $\frac{a+3b-c}{c}$= $\frac{-a+b+3c}{a}$= $\frac{a-b+3c}{b}$ tính P= (3 +$\frac{a}{b}$).(3+ $\frac{b}{c}$).(3+ $\frac{c}{a}$ )
By Josie
Đáp án:
Giải thích các bước giải:
`\frac{a+3b-c}{c}=\frac{-a+b+3c}{a}=\frac{c-b+3a}{b}`
`⇒ \frac{a+3b-c}{c}+1=\frac{-a+b+3c}{a}+1=\frac{c-b+3a}{b}+1`
`⇒\frac{a+3b}{c}=\frac{b+3c}{a}=\frac{c+3a}{b}`
Áp dụng T/C dãy tỉ số = nhau
`\frac{a+3b}{c}=\frac{b+3c}{a}=\frac{c+3a}{b}=\frac{a+3b+b+3c+c+3a}{c+a+b}=\frac{4(a+b+c)}{a+b+c}=4`
Do đó:`P=(3+\frac{a}{b})(3+\frac{b}{c})(3+\frac{c}{a})`
`⇒P=\frac{3b+a}{b}×\frac{3c+b}{c}×\frac{3a+c}{a}`
`⇒P=\frac{3b+a}{c}×\frac{3c+b}{a}×\frac{3a+c}{b}`
`⇒P=4×4×4`
`⇒P=64`
Vậy `P=64`