a^2 - 2ab - 3b^2 = 0

<=> a^2 - 3ab + ab - 3b^2 = 0

<=> a(a - 3b) + b(a - 3b) = 0

<=> (a - 3b)(a + b) = 0

=> a - 3b = 0 hoặc a + b = 0

=> a = 3b hoặc a = -b

+ Nếu a = 3b

A = (7a+2b)/(2a+b) + (9a-5b)/(2a-b)

A = (7.3b+2b)/(2.3b+b) + (9.3b-5b)/(2.3b-b)

A = 23b/7b + 22b/5b

A = 23/7 + 22/5 = 269/35

+ Nếu a = -b

A = (7a+2b)/(2a+b) + (9a-5b)/(2a-b)

A = (-7b+2b)/(-2b+b) + (-9b-5b)/(-2b-b)

A = -5b/-b + (-14b/-3b)

A = 5 + 14/3 = 29/3

a^2-2ab-3b^2=0

=>a^2-3ab+ab-3b^2=0

=>a(a-3b)+b(a-3b)=0

=>(a+b)(a-3b)=0

mà a,b khác 0 => a+b khác 0

=>a-3b=0

=>a=3b

Thay vào A ta được:

A=(7a+2b)/(2a+b)+(9a-5b)/(2a-b)

=(7.3b+2b)/(2.3b+b)+(9.3b-5b)/(2.3b-b)

=23b/7b+22b/5b=23/7+22/5=......

ta có:a-2ab-3b²=0

=>a²-3ab+ab-3b²=0

=>a(a-3b)+b(a-3b)=0

=>(a+b)(a-3b)=0

vìa,b khác 0=>a-3b=0

=>a=3b

thay vào A ta được:

A=(7.3b+2b)/(2.3b+b)+9=(9.3b-5b)/(2.3b-b)

  =23b/7b+22b/5b

  =23/7+22/5

  =269/35

Vậy A=269/35

Ta có: \(9a^2-b^2=0\Rightarrow\left(3a-b\right)\left(3a+b\right)=0\Rightarrow\left\{{}\begin{matrix}3a-b=0\\3a+b=0\end{matrix}\right.\)

\(9a^3-\dfrac{1}{3}b^3=\dfrac{1}{3}\left(27a^3-b^3\right)=\dfrac{1}{3}\left(3a-b\right)\left(9a^2+3ab+b^2\right)=\dfrac{1}{3}.0.\left(9a^2+3ab+b^2\right)=0\)

Ngoài http://olm.vn/hoi-dap/question/779981.html còn cách khác

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(9a^3+3a^2+c\right)\left(\frac{1}{9a}+\frac{1}{3}+c\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow A\le\text{∑}\frac{a\left(\frac{1}{9a}+\frac{1}{3}+c\right)}{\left(a+b+c\right)^2}=\text{∑}\left(\frac{1}{9}+\frac{a}{3}+ac\right)\)

\(=\frac{1}{3}+\frac{a+b+c}{3}+\text{∑}ab\le\frac{1}{3}+\frac{1}{3}+\frac{\left(a+b+c\right)^2}{3}=1\)

Dấu "=" khi \(a=b=c=\frac{1}{3}\)

a.b.c=1 thật hả. Rắc rối thế. Để nghĩ tiếp

Áp dụng BĐT AM-GM ta có:

\(9a^3+\frac{1}{3}+\frac{1}{3}\ge3\sqrt[3]{9a^3\cdot\frac{1}{3}\cdot\frac{1}{3}}=3a\)

\(3b^2+\frac{1}{3}\ge2\sqrt{3b^2\cdot\frac{1}{3}}=2b\)

Do đó: \(A\le\text{∑}\frac{a}{3a+2b+c-1}=\frac{a}{2a+b}\left(a+b+c=1\right)\)

\(2A\le\text{∑}\frac{2a}{2a+b}=3-\text{∑}\frac{b}{2a+b}=3-\text{∑}\frac{b^2}{2ab+b^2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(2A\le3-\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)

\(=3-\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=2\Leftrightarrow A\le1\)

Dấu "=" khi \(a=b=c=\frac{1}{3}\)

ĐK \(9a^2-b^2\ne0\)

Ta có B =\(\frac{2a-b}{3a-b}+\frac{5b-a}{3a+b}=\frac{\left(2a-b\right)\left(3a+b\right)+\left(5b-a\right)\left(3a-b\right)}{\left(3a+b\right)\left(3a-b\right)}\)

=\(\frac{6a^2+2ab-3ab-b^2+15ab-5b^2-3a^2+ab}{9a^2-b^2}\)

=\(\frac{3a^2+15ab-6b^2}{9a^2-b^2}=\frac{3\left(a^2+5ab-2b^2\right)}{9a^2-b^2}\)

Từ \(10a^2-3b^2+5ab=0\Rightarrow5ab=3b^2-10a^2\)

\(\Rightarrow B=\frac{3\left(a^2+3b^2-10a^2-2b^2\right)}{9a^2-b^2}=\frac{3\left(-9a^2+b^2\right)}{9a^2-b^2}=-3\)

Vậy B =-3

x²(y+z)+y²(z+y)+z²(x+y)