Question

Cho hai biểu thức :

\(M=\left(8x^6-27\right):\left(4x^4+6x^2+9\right)\)

\(N=\left(y^4-1\right):\left(y^3+y^2+y+1\right)\)

Tính tỉ số \(\frac{M}{N}\) với \(x=8,y=251\)

Answer 1

Ta có : \(M=\frac{8x^6-27}{4x^4+6x^2+9}=\frac{\left(2x^2\right)^3-3^3}{\left(2x^2\right)^2+\left(2x^2\right).3+3^2}\)

\(=\frac{\left(2x^2-3\right)\left[\left(2x^2\right)^2+2x^2.3+3^2\right]}{\left(2x^2\right)^2+2x^2.3+3^2}=2x^2-3\)

\(N=\frac{y^4-1}{y^3+y^2+y+1}=\frac{\left(y-1\right)\left(y^3+y^2+y+1\right)}{y^3+y^2+y+1}=y-1\)

Vậy \(\frac{M}{N}=\frac{2x^3-3}{y-1}\)

Khi \(x=8,y=251\) , ta có :

\(\frac{M}{N}=\frac{2.8^3-3}{251-1}=\frac{1}{2}\)