a, ĐKXĐ: x≠±2

A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)

A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)

A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)

b, |x|=\(\dfrac{1}{2}\)

TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)

TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)

Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:

\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)

\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)

c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)² < 0

⇔   {x-2>0        ⇔      {x>2

     [                           [

       {x+2<0                 {x<2

⇔   {x-2<0        ⇔      {x<2

     [                           [

       {x+2>0                 {x>2

⇔ x<2 

Vậy x<2 (trừ -2)

b) trước hết ta cần chứng minh nếu x+y+z=0 thì x^3+y^3+z^3=3xyz

ta có x+y+z=0==> x=-(y+z) 

             <=> \(x^3=-\left(y^3+z^3+3yz\left(y+z\right)\right)\)

           <=> \(x^3+y^3+z^3=-3yz\left(y+z\right)\)

      <=> \(x^3+y^3+z^3=3xyz\)( cì y+z=-x)

 áp dụng vào bài ta có \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

 do đó M=\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)

\(A=\left(b+c\right)^2+b^2+c^2=2b^2+2c^2+2bc=2\left(b^2+bc+c^2\right)\) (tự hiểu nhé)

Mà \(a^2=2\left(a+c+1\right)\left(a+b-1\right)=2a^2+2\left(ab+bc+ca\right)+2\left(b-c\right)-2\)

\(\Leftrightarrow a^2+2a\left(b+c\right)+2bc-2=0\) (*)

\(\Leftrightarrow2bc=2-a^2-2a\left(b+c\right)=2-\left(b+c\right)^2+2\left(b+c\right)^2\) (mấy cái này là từ a + b + c =0 suy ra a = -(b+c) suy ra a² = [-(b+c)]² = (b+c)² thôi!)

\(\Leftrightarrow\left(b+c\right)^2-2bc=-2\)

hay c² + b² = -2?? hay là mình làm sai nhì?

\(a^2=2\left(a+c+1\right)\left(a+b-1\right)\)

\(\Leftrightarrow\left(b+c\right)^2=\left(b-1\right)\left(c+1\right)\)

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

\(\Rightarrow a=0,b=1,c=-1\)

\(\Rightarrow A=2\)

Dòng số 2 bấm thiếu số 2 ở bên vế phải nha. B tự thêm số 2 vào nha. Còn lại thì không thay đổi gì nữa

Ta có \(\dfrac{a}{2009}\)=\(\dfrac{b}{2010}\)=\(\dfrac{c}{2011}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{c-a}{2011-2009}=\dfrac{c-a}{2}\left(1\right)\)

\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{a-b}{2009-2010}=\dfrac{a-b}{-1}\)(2)\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{b-c}{2010-2011}=\dfrac{b-c}{-1}\left(3\right)\)

Từ (1),(2),(3) \(_{\Rightarrow}\)\(\dfrac{c-a}{2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\Rightarrow\dfrac{\left(a-c\right)^{ }2}{2^{ }2}=\dfrac{\left(a-b\right)}{-1}\times\dfrac{\left(b-c\right)}{-1}\)

\(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(a-b\right)\times\left(b-c\right)}{1}\Rightarrow4\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\)

\(\Rightarrow M=4\left(a-b\right).\left(a-c\right)-\left(c-a\right)^2=0\)

Vậy M = 0

đặt \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=k\) ta có:

\(\Rightarrow a=2009k\left(1\right)\\ \Rightarrow b=2010k\left(2\right)\\ \Rightarrow c=2011k\left(3\right)\)

thay 1;2;3 vào M ta có:

\(M=4\left(2009k-2010k\right)\left(2010k-2011k\right)-\left(2011k-2009k\right)^2\\ \Rightarrow M=4.\left(-k\right)\left(-k\right)-\left(2k\right)^2\\ \Rightarrow M=4k^2-\left(2k\right)^2\\ \Rightarrow M=\left(2k\right)^2-\left(2k\right)^2\\ \Rightarrow M=0\)Vậy M = 0

\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)

Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)

\(=\dfrac{27}{2}\)

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

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào bt N được

\(N=\left(-13\right)\cdot\dfrac{1}{2}\cdot\left(-3\right)+2\cdot\dfrac{1}{2}+\left(-3\right)-2\)

\(=\dfrac{31}{2}\)

Vậy: Giá trị của N tại \(a=\dfrac{1}{2};b=-3\) là \(\dfrac{31}{2}\)

TH1: Nếu a+b+c \(\ne0\)

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

mà \(\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1=2\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=2\)

Vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=8\)

TH2 : Nếu a+b+c = 0

áp dụng tính chất của dãy tỉ số bằng nhau ta có :

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

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

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

vậy \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{b+c}{b}\right)=1\)

\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(\Leftrightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)

TH1: a+b+c=0 

\(\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\Rightarrow B=\left(1-\frac{a+c}{a}\right).\left(1-\frac{b+c}{c}\right).\left(1-\frac{a+b}{b}\right)=-1\)

TH2: a+b+c khác 0

 \(\Rightarrow a=b=c\Rightarrow B=\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right).\left(1+\frac{a}{a}\right)=2^3=8\)