Dạng 3. Tính giá trị của phân thức có đáp án

$\frac{-2x^2-40y+16x+5xy}{x^2+24y-8x-3xy}$ với $\frac{x}{y}=\frac{10}{3}$.

$\frac{-2x^2-40y+16x+5xy}{x^2+24y-8x-3xy}=\frac{-2x(x-8)-5y(8-x)}{x(x-8)-3y(x-8)}=\frac{(x-8)(5y-2x)}{(x-8)(x-3y)}=\frac{5y-2x}{x-3y}$

$P=\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}$$=\frac{a^2(a-4)-(a-4)}{a^3-2a^2-5a^2+10a+4a-8}$

$P=\frac{\left(a^2-1\right)(a-4)}{a^2(a-2)-5a(a-2)+4(a-2)}$$=\frac{(a+1)(a-1)(a-4)}{(a-1)(a-4)(a-2)}=\frac{a+1}{a-2}$

$B=\frac{x^3(x-1)-(x-1)}{x^2\left(x^2+x+1\right)+2\left(x^2+x+1\right)}$$=\frac{{(x-1)}^2\left(x^2+x+1\right)}{\left(x^2+2\right)\left(x^2+x+1\right)}=\frac{{(x-1)}^2}{x^2+2}$

$B=\frac{{(x-1)}^2}{x^2+2}\ge 0$

$B=\frac{1-x^4}{x^{10}-x^8+4x^6-4x^4+4x^2-4}$$=\frac{-\left(1-x^2\right)\left(1+x^2\right)}{\left(1-x^2\right)\left(x^8+4x^4+4\right)}$$=-\frac{\left(1+x^2\right)}{{\left(x^4+2\right)}^2}<0$ với $x\ne \pm 1.$