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

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)

1, Ta có a^3+b^3+c^3=3abc

-> a^3+b^3+c^3+3a^2b+3ab^2=3abc+3a^2b+3ab^2

-> (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^2+2ab+b^2-ac-bc+c^2-3ab)=0

Th1: a+b+c=0

->P= a+b/2 . b+c/2 . c+a/2

= (-c)(-a)(-b)/2=-1

TH2 a^2+b^2+c^2-ab-bc-ca=0

->2a^2+2b^2+2c^2-2ab-abc-2ac=0

->(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0

-> (a-b)^2+(a-c)^2+(b-c)^2=0

Mà (a-b)^2+(a-c)^2+(b-c)^2>= 0

Dấu = xảy ra (=)a-b=0

                         b-c=0

                          a-c=0

-> a=b=c

->P= 1+a/b+1+b/c+1+c/a=2+2+2= 8

bn có thể giải thích phần TH1 ko?

Phần í cx dễ hiểu mà

a: A\B={-3;-2} nên A={-3;-2;x}

B\A={6;9;10} nên B={6;9;10;y}

A giao B={0;1;2;3;4} nên A={-3;-2;0;1;2;4}; B={6;9;10;0;1;2;3;4}

b: A\B={4;5} nên A={4;5;x}

B\A={6;9} nen B={6;9;y}

A giao B={1;2;3} nên A={4;5;1;2;3}; B={6;9;1;3;2}

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

Lời giải:

Đặt \((ab,bc,ac)=(x,y,z)\)

Theo bài ra ta có:

\(x^3+y^3+z^3=3xyz\Leftrightarrow x^2+y^3+z^3-3xyz=0\)

\(\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\)

TH1:

\(x+y+z=0\) \(\Leftrightarrow ab+bc+ac=0\)

\(\Rightarrow M=\frac{1}{(a+b)(b+c)(c+a)}=\frac{1}{(a+b+c)(ab+bc+ac)-abc}=\frac{-1}{abc}\)

TH2:

\(x^2+y^2+z^2=xy+yz+xz\)

Theo BĐT AM-GM ta luôn có \(x^2+y^2+z^2\geq xy+yz+xz\)

Dấu bằng xảy ra khi

\(x=y=z\Leftrightarrow ab=bc=ac\Leftrightarrow a=b=c\)

Khi đó, \(M=\frac{1}{(a+b)(b+c)(c+a)}=\frac{1}{2a.2b.2c}=\frac{1}{8abc}\)

a: \(3x-\left|2x+1\right|=2\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2\right)^2-\left(2x+1\right)^2=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2-2x-1\right)\left(3x-2+2x+1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(5x-1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow x=3\)

e: Ta có: \(2n-3⋮n+1\)

\(\Leftrightarrow2n+2-5⋮n+1\)

\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{0;-2;4;-6\right\}\)