Bài giải

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\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{a^2\left(b-c\right)-b^2\left(a-c\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{a^2b-a^2c-b^2a+b^2c}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(c-b\right)\left(c-a\right)}\)

\(P=\frac{1}{a-b}.\frac{\left(a-b\right)\left(ab-ac-bc\right)}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(b-c\right)\left(a-c\right)}\)

\(P=\frac{ab-ac-bc}{\left(a-c\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)

\(P=\frac{ab-ac-bc+c^2}{\left(a-c\right)\left(b-c\right)}=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(a-c\right)\left(b-c\right)}=\frac{\left(a-c\right)\left(b-c\right)}{\left(a-c\right)\left(b-c\right)}\)

=> P = 1

Đáp số: P=1

\(P=-\frac{a^2}{\left(a-b\right)\left(c-a\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}-\frac{c^2}{\left(c-a\right)\left(b-c\right)}\)

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

\(=-\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=1\)

a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)

Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)

b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)

\(=\left(a-b\right)\left(a-c\right)\)

Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)

Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)

Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)

a) Ta có : \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\Leftrightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{c+a}{b}+1\)

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

\(\Rightarrow P=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

b) Đề bài sai ^^

với ab+bc+ca=1 

=>\(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)

tương tự mấy cái kia rồi thay vào, ta có

A=\(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)

b),ta có \(a^2+2bc-1=a^2+bc-ab-ac=\left(a-b\right)\left(a-c\right)\)

tương tự mấy cái kia, rồi thay váo, ta có 

\(B=\frac{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=1\)

^_^

Ta có:   MS = (1+a²).(1+b²).(1+c²)

= (ab + ac + bc + a²).(ab + ac + bc + b²).(ab + bc + ac + c²)

= [ (a² + ac) + (ab + bc) ] . [ (ab + b²) + (ac + bc) ] . [ (ab + bc) + (ac + c²) ]

= [ a(a + c) + b(a + c) ] . [ b(a + b) + c(a + b) ] . [ b(a + c) + c(a + c) ]

= (a + b)(a + c)(b + c)(a + b)(b + c)(a + c)

= (a + b)²(b + c)²(a + c)²     =  TS

Vậy   A = 1

mih tưởng câu B ra -1 chứ

@vũ tiền châu