Question

Cho \(a,b,c\ne0\) và \(a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}=\frac{c+2a-b}{b}\)

Tính giá trị biểu thức: \(P=\left(2+\frac{a}{b}\right).\left(2+\frac{b}{c}\right).\left(2+\frac{c}{a}\right)\)

Answer 1

Xét \(a+b+c=0\) thì \(\hept{\begin{cases}a+2b=c\\b+2c=a\\c+2a=b\end{cases}}\)\(\Rightarrow P=\frac{\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)}{abc}=1\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(a+b+c=\frac{a+2b-c}{c}=\frac{b+2c-a}{a}+\frac{c+2a-b}{b}=\frac{a+2b-c+b+2c-a+c+2a-b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=2\)

\(\Rightarrow\hept{\begin{cases}a+2b=3c\\b+2c=3a\\c+2a=3b\end{cases}}\)\(\Rightarrow P=\frac{3a.3b.3c}{abc}=27\)

Answer 2

Có a+2b-c/c=b+2c-a/a=c+2a-b/b

suy ra a+2b-c/c=b+2c-a/a=c+2a-b/b=a+2b-c+b+2c-a+c+2a-b/a+b+c=2a+2b+2c/a+b+c=2

suy ra a+2b-c=2c suy ra a+2b=3c

           b+2c-a=2a suy ra b+2c=3a

           c+2a-b=2b suy ra c+2a=3b

Có P=(2+a/b)(2+b/c)(2+c/a)=(2b+a/b)(2c+b/c)(2a+c/a)=(3c/b)(3a/c)(3b/a)=27abc/abc=27

Answer 3

Đồ con lợn