cho \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
voi \(x\ne\pm2\)
a) rut gon A
b)tim x de \(A-\frac{1}{x}=-1\)
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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~~~
\(a,x^2-3x=0\)
\(\Rightarrow x\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
- 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 .
a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)
\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)