a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)

\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)

\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)

giúp mik vs

ta có: (a^2-a+2012)(a^2-a+2014)-3

=(a^2-a+2013-1)(a^2-a+2013+1)-3

=(a^2-a+2013)^2-1-3

=(a^2-a+2013)^2-4

=(a^2-a+2013-2)(a^2-a+2013+2)

=(a^2-a+2011)(a^2-a+2015)

chúc bn học tốt

24 + 25 + 12 + 2018 = 2079

Mik cũng chúc bn giáng sinh vui vẻ bên gia đình, người thân và bạn bè ~Merry Christmas~

~~~Leo~~~

=2079 

k nhak

Love

\(a,x^2-3x=0\)

\(\Rightarrow x\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

- Thay \(x=0\) vào biểu thức A, ta được :

\(\frac{0-5}{0-4}=\frac{-5}{-4}=\frac{5}{4}\)

- Thay \(x=3\) vào biểu thức A, ta được :  

\(\frac{3-5}{3-4}=\frac{-2}{-1}=2\)

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

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

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

\(=\frac{x^2-25+2x^2-12x-2x^2+2x+50}{2x\left(x-5\right)}\)

\(=\frac{x^2-10x+25}{2x\left(x-5\right)}=\frac{\left(x-5\right)^2}{2x\left(x-5\right)}=\frac{x-5}{2x}\)

\(a,x^3+y^3+z^3-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz\right)-3xy\left(x+y+z\right)\)

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

b) \(f\left(x\right)=12x^3-32x^2+25x-6\)

Thấy \(x=\frac{3}{2}\) là một nghiệm.Vậy đa thức có chứa nhân tử \(\left(x-\frac{3}{2}\right)\)

Ta có: \(f\left(x\right)=\left(x-\frac{3}{2}\right)\left(\frac{12x^3-32x^2+25x-6}{x-\frac{3}{2}}\right)\)

\(=\left(x-\frac{3}{2}\right)\left(12x^2-14x+4\right)\)

\(=\left(x-\frac{3}{2}\right)\left[\left(12x^2-6x\right)-\left(8x-4\right)\right]\)

\(=\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(12x-6\right)\)

\(=12\left(x-\frac{3}{2}\right)\left(x-\frac{2}{3}\right)\left(x-\frac{1}{2}\right)\)

Mk làm chi tiết từng bc một nên hơi dài 

~ học tốt ~

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

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

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

\(=\frac{x^2\left(x-2\right)-\left(2x^2-4x+3x-6\right)}{x-2}\)

\(=\frac{x^2\left(x-2\right)-[\left(2x^2-4x\right)+\left(3x-6\right)]}{x-2}\)

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

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

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

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

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

\(=\frac{\left(x-2\right)[x\left(x-1\right)-3\left(x-1\right)}{x-2}\)

\(=\frac{\left(x-2\right)\left(x-1\right)\left(x-3\right)}{x-2}\)

\(=\left(x-1\right)\left(x-3\right)\)

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

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

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

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

\(=x^2-2x-3\)

@TrầnMinhPhong.

Đến đoạn này là được rồi . 

sai đề rồi bạn ơi

Bài nài cách làm đơn giản ạ :>>>

\(x^2+5x-6\)

\(=x^2-1x+6x-6\)

\(=\left(x^2-1x\right)+\left(6x-6\right)\)

\(=x\left(x-1\right)+6\left(x-1\right)\)

\(=\left(x-1\right)\left(x+6\right)\)

Hok tốt