Cho $\text{H}=\frac{15}{2}-\left(\frac{3}{5}+\frac{7}{2}\right)+\frac{-2}{5};\text{\hspace{0.17em}\hspace{0.17em}G = }\frac{3}{4}+\left(\frac{-7}{15}-\frac{-1}{4}\right)+\frac{8}{-15}$ . Khẳng định nào sau đây là sai

Đáp án đúng là: D

$\text{H}=\frac{15}{2}-\left(\frac{3}{5}+\frac{7}{2}\right)+\frac{-2}{5}=\frac{15}{2}-\frac{3}{5}-\frac{7}{2}+\frac{-2}{5}=\left(\frac{15}{2}-\frac{7}{2}\right)+\left(\frac{-3}{5}+\frac{-2}{5}\right)=4-1=3$.

$\text{G = }\frac{3}{4}+\left(\frac{-7}{15}-\frac{-1}{4}\right)+\frac{8}{-15}=\frac{3}{4}+\frac{-7}{15}-\frac{-1}{4}+\frac{-8}{15}=\left(\frac{3}{4}-\frac{-1}{4}\right)+\left(\frac{-7}{15}+\frac{-8}{15}\right)=1-1=0$.

Vì 3 > 0 nên H > G.

Vậy H < G là sai.