Ta thấy rằng:

\(a=\frac{1}{a}\times a\)

\(b=\frac{1}{b}\times b\)

\(c=\frac{1}{c}\times c\)

Suy ra: Kết quả cần tìm của a, b, c là abc. Ta có:

\(a\times b\times c=abc\)

Vậy:

\(abc=abc\times\frac{1}{a}\times\frac{1}{b}\times\frac{1}{c}\)

\(abc=\frac{a}{a}\times\frac{b}{b}\times\frac{c}{c}\)

\(abc=1\)

Vậy abc = 1.

Gía trị của biểu thức đó là:

      \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}.\)

\(=1+1+1+1+1+1.\)

\(=6\)

Vậy giá trị của biểu thức đó là 6.

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 bn tran mai phuong ha bn noi bn ay ngu chac bn ko ngu a

Có: \(\frac{3a+b+2c}{2a+c}=\frac{a+3b+c}{2b}=\frac{a+2b+2c}{b+c}\)

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

\(\Rightarrow\frac{a+b+c}{2a+c}+1=\frac{a+b+c}{2b}+1=\frac{a+b+c}{b+c}+1\)

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

\(\Rightarrow2a+c=2b=b+c\)

\(\Rightarrow\hept{\begin{cases}c=b\\a=\frac{1}{2}b\end{cases}}\)

Thay vào biểu thức trên , ta được:

\(P=\)\(\frac{\left(\frac{1}{2}b+b\right)\left(b+b\right)\left(b+\frac{1}{2}b\right)}{\frac{1}{2}b.b.b}=9\)

Vậy \(P=9\)

a)

xét ΔAHOΔAHO và ΔBHOΔBHO có:

OH(chung)

AHOˆ=BHOˆ=90oAHO^=BHO^=90o

O1ˆ=O2ˆ(gt)O1^=O2^(gt)

⇒ΔAHO=ΔBHO(g.c.g)⇒ΔAHO=ΔBHO(g.c.g)

=> OA=OB

b)

xét ΔACOΔACO và ΔBCOΔBCO có:

OA=OB(theo câu a)

O1ˆ=O2ˆO1^=O2^(gt)

OC(chung)

=>ΔACO=ΔABO(c.g.c)ΔACO=ΔABO(c.g.c)

=>{OACˆ=OBCˆCA=CB

a)

xét ΔAHOΔAHO và ΔBHOΔBHO có:

OH(chung)

AHOˆ=BHOˆ=90oAHO^=BHO^=90o

O1ˆ=O2ˆ(gt)O1^=O2^(gt)

⇒ΔAHO=ΔBHO(g.c.g)⇒ΔAHO=ΔBHO(g.c.g)

=> OA=OB

b)

xét ΔACOΔACO và ΔBCOΔBCO có:

OA=OB(theo câu a)

O1ˆ=O2ˆO1^=O2^(gt)

OC(chung)

=>ΔACO=ΔABO(c.g.c)ΔACO=ΔABO(c.g.c)

=>{OACˆ=OBCˆCA=CB