Question

[WonMaengGun]

Rút gọn biểu thức:

C= \(\frac{\sqrt{y^3}-1}{y+\sqrt{y}+1}-\frac{y+3\sqrt{y}+2}{\sqrt{y}+1}\:\) ( với y >0 hoặc y=0)

Answer 1

[HT.Phong (9A5)]

\(C=\dfrac{\sqrt{y^3}-1}{y+\sqrt{y}+1}-\dfrac{y+3\sqrt{y}+2}{\sqrt{y}+1}\) 

\(C=\dfrac{\sqrt{y^3}-1^3}{y+\sqrt{y}+1}-\dfrac{y+\sqrt{y}+2\sqrt{y}+2}{\sqrt{y}+1}\)

\(C=\dfrac{\left(\sqrt{y}+1\right)\left[\left(\sqrt{y}\right)^2+\sqrt{y}\cdot1+1\right]}{y+\sqrt{y}+1}-\dfrac{\left(\sqrt{y}+1\right)\left(\sqrt{y}+2\right)}{\sqrt{y}+1}\)

\(C=\dfrac{\left(\sqrt{y}+1\right)\left(y+\sqrt{y}+1\right)}{y+\sqrt{y}+1}-\left(\sqrt{y}+2\right)\)

\(C=\sqrt{y}+1-\sqrt{y}-2\)

\(C=-3\)

Answer 2

[Nguyễn Lê Phước Thịnh]

\(C=\dfrac{\left(\sqrt{y}-1\right)\left(y+\sqrt{y}+1\right)}{y+\sqrt{y}+1}-\dfrac{\left(\sqrt{y}+1\right)\left(\sqrt{y}+2\right)}{\sqrt{y}+1}\)

\(=\sqrt{y}-1-\sqrt{y}-2=-3\)