Câu 75:

Chứng minh các đẳng thức sau:

a) \(\left(\frac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right) \cdot \frac{1}{\sqrt{6}}=-1,5\)

b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \frac{1}{\sqrt{7}-\sqrt{5}}=-2\)

c) \(\frac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \frac{1}{\sqrt{a}-\sqrt{b}}=\mathbf{a}-\mathbf{b}\) với \(\mathrm{a}, \mathrm{b}\) dương và \(\mathrm{a} \neq \mathrm{b}\).

d) \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right) \cdot\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1\)-a với \(\mathrm{a} \geq 0\) và \(\mathrm{a} \neq 1\)

Lời giải:

Biến đổi vế trái:

a) \(\mathrm{VT}=\left(\frac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right) \frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2^{2} 3}-\sqrt{6}}{\sqrt{4.2}-2}-\frac{\sqrt{36.6}}{3}\right) \frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2} \sqrt{6}-\sqrt{6}}{2 \sqrt{2}-2}-\frac{6 \sqrt{6}}{3}\right) \frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}(\sqrt{2}-1)}{2(\sqrt{2}-1)}-2 \sqrt{6}\right) \frac{1}{\sqrt{6}}=\left(\frac{\sqrt{6}}{2}-2 \sqrt{2}\right) \cdot \frac{1}{\sqrt{6}}\)

\(=\frac{-3 \sqrt{6}}{2} \cdot \frac{1}{\sqrt{6}}=-\frac{3}{2}=-1,5=\operatorname{VP}(\mathrm{đpcm})\)

b) \(\mathrm{VT}=\left[\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right]: \frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{\sqrt{7} \sqrt{2}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{5} \sqrt{3}-\sqrt{5}}{1-\sqrt{3}}\right]: \frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{1-\sqrt{2}}+\frac{\sqrt{5}(\sqrt{3}-1)}{1-\sqrt{3}}\right]:(\sqrt{7}-\sqrt{5})\)

\(=(-\sqrt{7}-\sqrt{5})(\sqrt{7}-\sqrt{5})\)

\(=-(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})\)

\(=-(7-5)=-2=\operatorname{VP}\) (đpcm)

c) \(\mathrm{VP}=\frac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a^{3}} \cdot \sqrt{b}+\sqrt{b^{2}} \sqrt{a}}{\sqrt{a b}}: \frac{1}{\sqrt{a}-\sqrt{b}}\)

\(=\frac{\sqrt{a b}(\sqrt{a}+\sqrt{b})}{\sqrt{a b}} \cdot(\sqrt{a}-\sqrt{b})=(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\)

\(=a-b=VP\) (đpcm)

d) \(\mathrm{VT}=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left(1+\frac{\sqrt{a^{2}}+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a^{2}}-\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left[1+\frac{\sqrt{a}(\sqrt{a}+1)}{\sqrt{a}+1}\right]\left[1-\frac{\sqrt{a}(\sqrt{a}-1)}{\sqrt{a}-1}\right]\)

\(=(1+\sqrt{a})(1-\sqrt{a})\)

\(=1-(\sqrt{a})^{2}=1-a=V P\) (đpcm)