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- A =\(\frac{x^2+3+2x-6-x-3}{x^2-9}\)
- A =\(\frac{x^2+x-6}{x^2-9}\)
- A = \(\frac{\left(x+3\right)\left(x-2\right)}{\left(x+3\right)\left(x-3\right)}\)
- A = \(\frac{x-2}{x-3}\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
Ta có : A=20+21+22+23+...+22010A=20+21+22+23+...+22010
3A=2+22+23+24+...+220113A=2+22+23+24+...+22011
=> 2A=3A−A=(21+22+...+22011)−(20+21+...+22010)
=>2A=22011−12A=22011−1
=>A=22011−12A=22011−12
=> A < B ( vì 22011−12<2201122011−12<22011 )
a: \(=\dfrac{x}{y\left(x-y\right)}+\dfrac{2x-y}{y\left(x-y\right)}=\dfrac{x+2x-y}{y\left(x-y\right)}=\dfrac{3x-y}{y\left(x-y\right)}\)
b: \(=\dfrac{x\left(x+3\right)}{\left(x+3\right)^2}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x-3}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
c: \(=\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{x^2+9x-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x+3}{x-3}\)
d: \(=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)