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

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

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

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

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

Bài 2:

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

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

\(A=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)\)

\(A=\dfrac{1}{abc}\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]\)

Vì \(a+b+c=0\)

Nên a + b = -c (1)

Thay (1) vào A, ta được:

\(A=\dfrac{1}{abc}\left[\left(-c\right)^3-3ab\left(-c\right)+c^3\right]\)

\(A=\dfrac{1}{abc}.3abc\)

\(A=3\)

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

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

Vì \(a+b+c=0\)

Nên b + c = -a

=> ( b + c )² = (-a)²

=> b² + c² + 2bc = a²

=> b² + c² = a² - 2bc (1)

Tương tự ta có: c² + a² = b² - 2ac (2)

a² + b² = c - 2ab (3)

Thay (1), (2) và (3) vào B, ta được:

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

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

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

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

\(B=\dfrac{1}{2abc}\left(a^3+b^3+c^3\right)\)

Mà \(a^3+b^3+c^3=3abc\) ( câu a )

\(\Rightarrow B=\dfrac{1}{2abc}.3abc\)

\(\Rightarrow B=\dfrac{3}{2}\)

Bài 1:

a) GT: abc = 2

\(M=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{abc+2cb+2b}\)

\(M=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2+2cb+2b}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{2cb}{2\left(1+cb+b\right)}\)

\(M=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(M=\dfrac{1+b+bc}{bc+b+1}\)

\(M=1\)

b) GT: abc = 1

\(N=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(N=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{cb}{b\left(ac+c+1\right)}\)

\(N=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{bc}{abc+bc+b}\)

\(N=\dfrac{1}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{bc}{bc+b+1}\)

\(N=\dfrac{1+b+bc}{bc+b+1}\)

\(N=1\)

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

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

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

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

\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)

\(P\le\dfrac{a}{2\sqrt{a^2bc}}+\dfrac{b}{2\sqrt{b^2ca}}+\dfrac{c}{2\sqrt{c^2ab}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\right)\)

\(P\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{2}\left(\dfrac{ab+bc+ca}{abc}\right)\le\dfrac{1}{2}\left(\dfrac{a^2+b^2+c^2}{abc}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=3\)

Áp dụng cosi:

`a^2+bc>=2a\sqrt{bc}`

Hoàn toàn tương tự:

`=>P<=1/2(1/sqrt{ab}+1/sqrt{bc}+1/sqrt{ca})`

Áp dụng cosi:

`1/a+1/b+1/c>=1/sqrt(ab)+1/sqrt(bc)+1/sqrt(ca)`

`=>P<=1/2(1/a+1/b+1/c)`

`=>P<=1/2((ab+bc+ca)/(abc))<=(a^2+b^2+c^2)/(2(abc))=1/2`

Dấu "=" `<=>a=b=c=3`

\(P\ge\dfrac{3abc}{2abc}+\dfrac{a^2+b^2}{c^2+\dfrac{a^2+b^2}{2}}+\dfrac{b^2+c^2}{a^2+\dfrac{b^2+c^2}{2}}+\dfrac{c^2+a^2}{b^2+\dfrac{c^2+a^2}{2}}\)

\(P\ge\dfrac{3}{2}+2\left(\dfrac{a^2+b^2}{a^2+c^2+b^2+c^2}+\dfrac{b^2+c^2}{a^2+b^2+a^2+c^2}+\dfrac{a^2+c^2}{a^2+b^2+b^2+c^2}\right)\)

Đặt \(\left(a^2+b^2;b^2+c^2;a^2+c^2\right)=\left(x;y;z\right)\)

\(\Rightarrow P\ge\dfrac{3}{2}+2\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{3}{2}+2\left(\dfrac{x^2}{xy+xz}+\dfrac{y^2}{yz+xy}+\dfrac{z^2}{xz+yz}\right)\)

\(P\ge\dfrac{3}{2}+\dfrac{2\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3}{2}+\dfrac{3\left(xy+yz+zx\right)}{xy+yz+zx}=3+\dfrac{3}{2}=\dfrac{9}{2}\)

Dấu "=" xảy ra khi \(a=b=c\)

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

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

\(=2\left(a+b+c\right)\)