Toán Lớp 8: A/ $\frac{1}{x}$ – $\frac{1}{x+1}$
B/ $\frac{1}{xy-x^2}$ – $\frac{1}{y^2-xy}$
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 8: A/ $\frac{1}{x}$ – $\frac{1}{x+1}$
B/ $\frac{1}{xy-x^2}$ – $\frac{1}{y^2-xy}$
Comments ( 2 )
$\\$
a,
1/x-1/(x+1)(x\ne-1;x\ne0)
=(x+1)/(x(x+1))-x/(x(x+1))
=(x+1-x)/(x(x+1))
=1/(x(x+1))
$\\$
b,
1/(xy-x^2)-1/(y^2-xy)(x\ne0;y\ne0)
=1/(x(y-x))-1/(y(y-x))
=y/(xy(y-x))-x/(xy(y-x))
=(y-x)/(xy(y-x))
=1/(xy)
$a)$ $\dfrac{1}{x}-\dfrac{1}{x+1}$ (đk: x!=-1; x!=0)
$=\dfrac{x+1}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}$
$=\dfrac{x+1-x}{x\left(x+1\right)}$
$=\dfrac{1}{x\left(x+1\right)}$
__________________________________
$b)$ $\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}$ (đk:x!=0 ; y!=0 ; x^2 – x y!=0 ; x y – y^2!=0)
$=\dfrac{y^2-xy}{\left(-x^2+xy\right)\left(y^2-xy\right)}-\dfrac{-x^2+xy}{\left(-x^2+xy\right)\left(y^2-xy\right)}$
$=\dfrac{y^2-xy-\left(-x^2+xy\right)}{\left(-x^2+xy\right)\left(y^2-xy\right)}$
$=\dfrac{x^2+y^2-2xy}{\left(-x^2+xy\right)\left(y^2-xy\right)}$
$=\dfrac{1}{xy}$