Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

1. (a+b)^2 ≥ 4ab

<=> a²+2ab+b²≥ 4ab

<=> a²+2ab+b²-4ab≥ 0

<=> a²-2ab+b²≥ 0

<=> (a-b)^2 ≥ 0 ( luôn đúng )

2. a^2 + b^2 + c^2 ≥ ab + bc + ca

<=> 2a^2 + 2b^2 + 2c^2 ≥ 2ab + 2bc + 2ca

<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0

<=> (a^2- 2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2) ≥ 0

<=> (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0 ( luôn đúng)

2)  \(x^3-6x^2+11x-6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-3x^2+9x+2x-6=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-3x+2\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x-1\right)=0\)

bn giải tiếp nha

3)   \(x^3-4x^2+x+6=0\)

\(\Leftrightarrow\)\(x^3-3x^2-x^2+3x-2x+6=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x^2-x-2\right)=0\)

\(\Leftrightarrow\)\(\left(x-3\right)\left(x-2\right)\left(x+1\right)=0\)

lm tiếp nha

4)  \(x^3-3x^2+4=0\)

\(\Leftrightarrow\)\(x^3+x^2-4x^2-4x+4x+4=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\)\( \left(x+1\right)\left(x-2\right)^2=0\)

lm tiếp nha

Mk làm mẫu 1 bài cho nha !

1. <=> (x^3-x^2)+(5x^2-5x)+(6x-6) = 0

<=> (x-1).(x^2+5x+6) = 0

<=> (x-1).[(x^2+2x)+(3x+6)] = 0

<=> (x-1).(x+2).(x+3) = 0

<=> x-1=0 hoặc x+2=0 hoặc x+3=0

<=> x=1 hoặc x=-2 hoặc x=-3

Vậy ..............

Tk mk nha

Lời giải:
\(P=\frac{3}{ab+bc+ac}+\frac{5}{(a+b+c)^2-2(ab+bc+ac)}=\frac{3}{ab+bc+ac}+\frac{5}{1-2(ab+bc+ac)}\)

\(=\frac{3}{x}+\frac{5}{1-2x}\) với $x=ab+bc+ac$

Theo BĐT AM-GM:
$1=(a+b+c)^2\geq 3(ab+bc+ac)$

$\Rightarrow x=ab+bc+ac\leq \frac{1}{3}$

Vậy ta cần tìm min $P=\frac{3}{x}+\frac{5}{1-2x}$ với $0< x\leq \frac{1}{3}$

Áp dụng BĐT Bunhiacopxky:

$(\frac{3}{x}+\frac{5}{1-2x})[2x+(1-2x)]\geq (\sqrt{6}+\sqrt{5})^2$

$\Leftrightarrow P\geq (\sqrt{6}+\sqrt{5})^2=11+2\sqrt{30}$

Vậy $P_{\min}=11+2\sqrt{30}$

Giá trị này đạt tại $x=3-\sqrt{\frac{15}{2}}$

Con cảm ơn cô ạ