\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)

\(\Rightarrow VT=\dfrac{ab}{a^2+b^2-c^{^2}}+\dfrac{bc}{b^2+c^2-a^{^2}}+\dfrac{ca}{c^2+a^2-b^{^2}}\\ =\dfrac{ab}{a^2+\left(b+c\right)\left(b-c\right)}+\dfrac{bc}{b^2+\left(c+a\right)\left(c-a\right)}+\dfrac{ca}{c^2+\left(a+b\right)\left(a-b\right)}\\ =\dfrac{ab}{a^2-a\left(b-c\right)}+\dfrac{bc}{b^2-b\left(c-a\right)}+\dfrac{ca}{c^2-c\left(a-b\right)}\\ =\dfrac{b}{a-b+c}+\dfrac{c}{b-c+a}+\dfrac{a}{c-a+b}\\ =\dfrac{b}{\left(a+c\right)-b}+\dfrac{c}{\left(a+b\right)-c}+\dfrac{a}{\left(c+b\right)-a}\\ =\dfrac{b}{-b-b}+\dfrac{c}{-c-c}+\dfrac{a}{-a-a}\\ =\dfrac{b}{-2b}+\dfrac{c}{-2c}+\dfrac{a}{-2a}\\ =-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}=-\dfrac{3}{2}=VP\)

a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{a^3+b^3+c^3}{abc}\)

\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)

b)Ta có: a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)

\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)

\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)

\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)

cm:nếu a+b+c=0 thì a^3+b^3+c^3=3abc

a^3+b^3+c^3=3abc

=>a^3+b^3+c^3-3abc=0

=>(a+b)^3-3ab(a+b)+c^3-3abc=0

=>[(a+b)^3+c^3]-3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2] -3ab(a+b+c)=0

=>(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]=0

vì a+b+c=0 nên a^3+b^3+c^3=3abc

thay kết quả vừa chúng minh vào đề bài ta đc

\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

chúc bạn học tốt ^ ^

Bài 1:

Từ \(a+b+c=0\) ta có:

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

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

\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)

Lại có:

\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3\)

\(=-c^3+3abc+c^3=3abc\)

Do đó \(B=\frac{3abc}{2abc}=\frac{3}{2}\)

Bài 2:

Lấy P-Q ta có:

\(P-Q=\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)\)

\(P-Q=\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{b^2+bc+c^2}+\frac{c^3-a^3}{c^2+ac+a^2}\)

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

\(P-Q=(a-b)+(b-c)+(c-a)=0\Rightarrow P=Q\)

Ta có đpcm.

ta chỉ cần cho a, b, c, thỏa mãn điều kiện

ví dụ ta cho a =1 ,b = -2 , c = 1

sau đó gán a, b, c đã thỏa mã điều kiện cho vào biểu thức

thì ta sẽ đươc kết quả giá trị của biểu thức là \(\dfrac{-5}{6}\)

vậy P =\(\dfrac{-5}{6}\)

Theo bài ra , ta có :

a+b+c = 0

(=) a+b = -c

(=) (a+b)² = c²

(=) a² + 2ab + b² = c²

(=) a² + b² - c² = -2ab

CMTT , ta có :

b² + c² - a² = -2bc

c² + a² - b² = -2ac

=) P = -1/2 + -1/2 + -1/2 = -3/2

âm 3/2 nha

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)

=> ac + bc = ab + ac = bc + ab (do abc \(\ne0\))

=> ac + bc - ab - ac = 0

=> bc - ab = 0

=> b(c - a) = 0

Mà b \(\ne0\) nên c - a = 0 => c = a

Tương tự ta có: a = b

Từ đó có: a = b = c

Thay vào M được:

\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Do \(a,b,c\ne0\)

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)

Ta có:

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\)

<=> \(ab\cdot\left(b+c\right)=bc\cdot\left(a+b\right)\)

<=> \(b^2\cdot\left(a-c\right)=0\)

<=> \(a=c\)

Làm tương tự ta được \(b=a\) => a=b=c

=> M=1