Giải bài 37 trang 67

Đề bài

a) Cho biểu thức: $C=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{24}}+\frac{1}{\sqrt{25}}.$ Chứng minh $C>\frac{24}{5}.$

b) Cho biểu thức $D=\left(\frac{y-2}{y+2\sqrt{y}}+\frac{1}{\sqrt{y}+2}\right).\frac{\sqrt{y}+1}{\sqrt{y}-1}$ với $y>0,y\ne 1.$ Chứng minh $D=\frac{\sqrt{y}+1}{\sqrt{y}}.$

Lời giải chi tiết

a) Ta có: $2<3<4<...<25$ nên $\sqrt{2}<\sqrt{3}<\sqrt{4}<...<\sqrt{25}$, do đó $\frac{1}{\sqrt{2}}>\frac{1}{\sqrt{3}}>\frac{1}{\sqrt{4}}>...>\frac{1}{\sqrt{25}}$.

Suy ra $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{25}}>\frac{1}{\sqrt{25}}+\frac{1}{\sqrt{25}}+\frac{1}{\sqrt{25}}+...+\frac{1}{\sqrt{25}}$ (24 hạng tử $\frac{1}{\sqrt{25}}$).

Hay $C>24.\frac{1}{\sqrt{25}}$. Vậy $C>\frac{24}{5}$.

b) $D=\left(\frac{y-2}{y+2\sqrt{y}}+\frac{1}{\sqrt{y}+2}\right).\frac{\sqrt{y}+1}{\sqrt{y}-1}$

 $\begin{array}{c}=\left(\frac{y-2}{\sqrt{y}\left(\sqrt{y}+2\right)}+\frac{1}{\sqrt{y}+2}\right).\frac{\sqrt{y}+1}{\sqrt{y}-1}\\ =\frac{y-2+\sqrt{y}}{\sqrt{y}\left(\sqrt{y}+2\right)}.\frac{\sqrt{y}+1}{\sqrt{y}-1}\\ =\frac{\left(\sqrt{y}-1\right)\left(\sqrt{y}+2\right)}{\sqrt{y}\left(\sqrt{y}+2\right)}.\frac{\sqrt{y}+1}{\sqrt{y}-1}\\ =\frac{\sqrt{y}+1}{\sqrt{y}}\end{array}$

Vậy $D=\frac{\sqrt{y}+1}{\sqrt{y}}.$