\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

<=>  \(\frac{ab+bc+ca}{abc}=0\)

<=>  \(ab+bc+ca=0\)

=>  \(ab+bc=-ca\)

<=>  \(\left(ab+bc\right)^3=-ca^3\)

Ta co:   \(a^3b^3+b^3c^3+c^3a^3=a^3b^3+b^3c^3-\left(ab+bc\right)^3=a^3b^3+b^3c^3-ab^3-bc^3-3ab.bc\left(ab+bc\right)\)

\(=-3ab.bc\left(ab+bc\right)=-3ab.bc.\left(-ca\right)=3a^2b^2c^2\)

\(M=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{b^3c^3+c^3a^3+a^3b^3}{abc}=\frac{3a^2b^2c^2}{abc}=3abc\)

\(5\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)

\(B_{min}\Leftrightarrow2-5\sqrt{x}ĐạtGTLNvà:\sqrt{x}+3đạtGTNN\)

Dấu "=" xảy ra <=> x=0

Vậy B_min=2/3

Ghi nốt đề đi

\(A=\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

 \(=6x^2+33x-10x-55-6x^2-14x-9x-21\)

  \(=-73\)ko phụ thuộc vào biến x

Vậy

A = 6x +33x - 10x - 55 -6x -14x -9x - 21

   = 76 -> Ðpcm

Đề sai sửa luôn !

\(a,M=\left(\frac{21}{x^2-9}+\frac{4-x}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21-\left(4-x\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\frac{x+3-1}{x+3}\right)\)

\(=\frac{21-4x-12+x^2+3x-x^2+3x+x-3}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(=\frac{3x+6}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\)

\(=\frac{3}{x-3}\)

\(b,x^2-4=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Kết hợp ĐKXĐ => x = 2

Thay vào \(M=\frac{3}{2-3}=\frac{3}{-1}=-3\)

Vậy ...........................

biết đề ghê vậy :D ?!

\(\left(x^2+\frac{2}{5}y\right)\left(x^2+\frac{2}{5}y\right)=x^4+\frac{4}{5}x^2y+\frac{4}{25}y\)

b\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2-\frac{1}{3}\right)^3-x^2-\frac{1}{3}\)

=(x²-1/3)³-x²+1/3 nhé

VN cố lên VN vô địch ^_^ 

k nhé

U23 Việt Nam chắc chắn sẽ mãi mãi vô địch! Cố lên U23 Việt Nam vô địch