Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Rightarrow2+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Rightarrow2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=2\)

\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=1\)

\(\Rightarrow\frac{a+b+c}{abc}=1\Rightarrow a+b+c=abc\left(đpcm\right)\)

Bài 1:

a)

 \(9x^2-49=0\)

\(9x^2-49+49=0+49.\)

\(9x^2=49\)

\(\frac{9x^2}{9}=\frac{49}{9}\)

\(x^2=\frac{49}{9}\)

\(x=\sqrt{\frac{49}{9}}\)

\(x=\frac{\sqrt{49}}{\sqrt{9}}\)

\(x=\frac{7}{3}\)hay \(x=2,33333...\)

b)

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

\(x^2+x-2-x-2.\)

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

\(x^2-4=0\)

\(x=\sqrt{4}\)

\(x=2\)

Bài 2:

a)

      \(\frac{x}{x}-3+9-\frac{6x}{x^2}-3x.\)

\(=1-3+9-\frac{6x}{x^2}-3x.\)

\(=1-3+9-\frac{6}{x}-3x.\)

\(=7-\frac{6}{x}-3x\)

b)

       \(6x-\frac{3}{x}\div4x^2-\frac{1}{3x^2}\)

\(=6x-\frac{3}{x}\div\frac{4}{1}x^2-\frac{1}{3x^2}.\)

\(=6x-\frac{3}{x}\times\frac{1}{4}x^2-\frac{1}{3x^2}\)

\(=6x-\frac{3x^2}{x4}-\frac{1}{3x^2}\)

\(=6x-\frac{3x}{4}-\frac{1}{3x^2}\)

\(=\frac{6x}{1}-\frac{3x}{4}-\frac{1}{3x^2}\)

\(=\frac{72x^3-36x^3-12x^2}{12x^2}\)

\(=\frac{36-12x^2}{12x^2}\)

có thêm điều kiện gì ko ? 

ko có điều kiện gì tức là \(a^{2017}+b^{2017}\)có  đáp án =0 hoặc  =1 đấy

Xem họ cái ông ơi nhanh tui off

a) Ta có: A = \(\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+4x}{4-x^2}\)

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

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

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

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

b) Với x = 4 => A = \(\frac{4-2}{4+2}=\frac{2}{8}=\frac{1}{4}\)

c) ĐKXĐ: \(\hept{\begin{cases}x-2\ne0\\x+2\ne0\\4-x^2\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne2\\x\ne-2\\x\ne\pm2\end{cases}}\) <=> \(x\ne\pm2\)

Ta có: A = \(\frac{x-2}{x+2}=\frac{\left(x+2\right)-4}{x+2}=1-\frac{4}{x+2}\)

Để A  nhận giá trị nguyên dương <=> \(1-\frac{4}{x+2}\) nguyên dương

<=> \(-\frac{4}{x+2}\) nguyên dương <=> -4 \(⋮\)x + 2

 <=> x + 2 \(\in\)Ư(-4) = {1; -1; 2; -2; 4; -4}

Lập bảng: 

Vậy ....

\(B=\frac{x+5}{2x}-\frac{x-6}{5-x}-\frac{2x^2-2x-50}{2x^2-10x}\)

\(B=\frac{x+5}{2x}-\left(\frac{x-6}{5-x}\right)-\left(\frac{2x^2-2x-50}{2x^2-10x}\right)\)

\(B=\frac{-2x^4+30x^3-150x^2+250x}{-4x^4+40x^3-100x^2}\)

\(B=\frac{-2x^3+30x^2-150x+250}{-4x^3+40x^2-100x}\)

\(B=\frac{-x^3+15x^2-75x+125}{-2x^3+20x^2-50x}\)

\(B=\frac{\left(-x+5\right)\left(x-5\right)\left(x-5\right)}{2x\left(-x+5\right)\left(x-5\right)}\)

\(B=\frac{x-5}{2x}\)

Ta có: M = x² + 6y + 10 + y² - x

          M = ( x² - x + 1/4 ) + ( y² + 6y + 9) + 3/4

          M = ( x - 1/2)² + ( y + 3 )² + 3/4

- Vì ( x - 1/2 )² >= 0 với mọi x; ( y + 3 )² >= 0 với mọi y => M >= 3/4 với moi x,y.

Dấu = xra <=> x - 1/2 = 0 và y + 3 = 0

                  <=> x = 1/2 và y = -3.

Ta có: x + y = 9

           x + z = 15

           y + z = 12

=> x + y + x + z + y + z = 9 + 15 + 12

<=> 2( x + y + z ) = 36

<=> x + y + z = 18

=> x = 18 - 12 = 6

     y = 18 - 15 = 3

     z = 18 - 9 = 9

Từ \(x+y=9\), \(x+z=15\), \(y+z=12\)

\(\Rightarrow x+y+x+z+y+z=9+15+12\)\(\Rightarrow2\left(x+y+z\right)=36\)

\(\Rightarrow x+y+z=18\)

Ta có: \(x+y=9\)\(\Rightarrow z=18-9=9\)

         \(x+z=15\)\(\Rightarrow y=18-15=3\)

         \(y+z=12\)\(\Rightarrow x=18-12=6\)

Vậy \(x=6\), \(y=3\), \(z=9\)