Đáp án:

Giải thích các bước giải:

 a) `7\sqrt{\frac{1}{7}}+\frac{35}{6\sqrt{7}}-\frac{7}{3}.\sqrt{\frac{4}{7}}-\frac{1}{2}\sqrt{63}`

`=\sqrt{7}+\frac{5\sqrt{7}}{6}-\frac{2\sqrt{7}}{3}-\frac{3\sqrt{7}}{2}`

`=-\frac{\sqrt{7}}{3}`

b) `\frac{15}{\sqrt{5}}-6\sqrt{\frac{5}{9}}+5\sqrt{\frac{1}{5}}-\sqrt{80}`

`=3\sqrt{5}-2\sqrt{5}+\sqrt{5}-4\sqrt{5}`

`=-2\sqrt{5}`

Bài 2:

a) `A=\frac{3}{\sqrt{3}}-\frac{2}{\sqrt{3}+1}+\frac{3-\sqrt{3}}{\sqrt{3}-1}`

`A=\sqrt{3}+1-\sqrt{3}+\frac{\sqrt{3}.(\sqrt{3}-1)}{\sqrt{3}-1}`

`A=\sqrt{3}+1-\sqrt{3}+\sqrt{3}`

`A=\sqrt{3}+1`

b) `B=\frac{5}{\sqrt{5}}-\frac{4}{\sqrt{5}+1}+\frac{5-\sqrt{5}}{\sqrt{5}-1}`

`B=\sqrt{5}+1-\sqrt{5}+\sqrt{5}`

`B=\sqrt{5}+1`

c) `C=\frac{6}{\sqrt{6}}-\frac{5}{\sqrt{6}+1}+\frac{6-\sqrt{6}}{\sqrt{6}-1}`

`C=\sqrt{6}+1-\sqrt{6}+\sqrt{6}`

`C=\sqrt{6}+1`

Giải thích các bước giải:

Bài 1:

a/ $7\sqrt{\dfrac{1}{7}}+\dfrac{35}{6\sqrt{7}}-\dfrac{7}{3}.\sqrt{\dfrac{4}{7}}-\dfrac{1}{2}.\sqrt{63}$

$=\dfrac{7}{\sqrt{7}}+\dfrac{35}{6\sqrt{7}}-\dfrac{14}{3\sqrt{7}}-\dfrac{\sqrt{63}}{2}$

$=\dfrac{42+35-28-\sqrt{63}.3\sqrt{7}}{6\sqrt{7}}$

$=\dfrac{49-3\sqrt{63.7}}{6\sqrt{7}}$

$=\dfrac{49-3.\sqrt{441}}{6\sqrt{7}}$

$=\dfrac{49-3.21}{6\sqrt{7}}$

$=\dfrac{-14}{6\sqrt{7}}$

$=\dfrac{-7\sqrt{7}}{3.7}$

$=\dfrac{-7\sqrt{7}}{21}$

b/ $\dfrac{15}{\sqrt{5}}-6\sqrt{\dfrac{5}{9}}+5\sqrt{\dfrac{1}{5}}-\sqrt{80}$

$=\dfrac{15}{\sqrt{5}}-\dfrac{6\sqrt{5}}{3}+\dfrac{5}{\sqrt{5}}-\sqrt{80}$

$=\dfrac{45-30+15-3\sqrt{400}}{3\sqrt{5}}$

$=\dfrac{45-30+15-3.20}{3\sqrt{5}}$

$=\dfrac{-30}{3\sqrt{5}}$

$=\dfrac{-10}{\sqrt{5}}$

$=\dfrac{-10\sqrt{5}}{5}$

Bài 2:

a/ $A=\dfrac{3}{\sqrt{3}}-\dfrac{2}{\sqrt{3}+1}+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}$

$⇔ A=\dfrac{3(\sqrt{3}-1)(\sqrt{3}+1)-2\sqrt{3}(\sqrt{3}-1)+(3-\sqrt{3}).\sqrt{3}.(\sqrt{3}+1)}{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}$

$⇔ A=\dfrac{3(3-1)-2.3+2\sqrt{3}+(3-\sqrt{3})(3+\sqrt{3})}{\sqrt{3}(3-1)}$

$⇔ A=\dfrac{3.2-2.3+2\sqrt{3}+9-3}{2\sqrt{3}}$

$⇔ A=\dfrac{6+2\sqrt{3}}{2\sqrt{3}}$

$⇔ A=\dfrac{3+\sqrt{3}}{\sqrt{3}}$

$⇔ A=\dfrac{\sqrt{3}(\sqrt{3}+1}{\sqrt{3}}$

$⇔ A=\sqrt{3}+1$

b/ $B=\dfrac{5}{\sqrt{5}}-\dfrac{4}{\sqrt{5}+1}+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}$

$⇔ B=\dfrac{5(\sqrt{5}-1)(\sqrt{5}+1)-4\sqrt{5}(\sqrt{5}-1)+(5-\sqrt{5}).\sqrt{5}.(\sqrt{5}+1}{\sqrt{5}(\sqrt{5}-1)(\sqrt{5}+1)}$

$⇔ B=\dfrac{5.(5-1)-4.5+4\sqrt{5}+(5-\sqrt{5}).(5+\sqrt{5})}{\sqrt{5}(5-1)}$

$⇔ B=\dfrac{5.4-4.5+4\sqrt{5}+25-5}{4\sqrt{5}}$

$⇔ B=\dfrac{20+4\sqrt{5}}{4\sqrt{5}}$

$⇔ B=\dfrac{5+\sqrt{5}}{\sqrt{5}}$

$⇔ B=\dfrac{\sqrt{5}(\sqrt{5}+1)}{\sqrt{5}}$

$⇔ B=\sqrt{5}+1$

c/ $C=\dfrac{6}{\sqrt{6}}-\dfrac{5}{\sqrt{6}+1}+\dfrac{6-\sqrt{6}}{\sqrt{6}-1}$

$⇔ C=\dfrac{6(\sqrt{6}-1)(\sqrt{6}+1)-5\sqrt{6}(\sqrt{6}-1)+(6-\sqrt{6}).\sqrt{6}(\sqrt{6}+1)}{\sqrt{6}(\sqrt{6}-1)(\sqrt{6}+1)}$

$⇔ C=\dfrac{6.(6-1)-5.6+5\sqrt{6}+(6-\sqrt{6})(6+\sqrt{6})}{\sqrt{6}(6-1)}$

$⇔ C=\dfrac{6.5-5.6+5\sqrt{6}+36-6}{5\sqrt{6}}$

$⇔ C=\dfrac{30+5\sqrt{6}}{5\sqrt{6}}$

$⇔ C=\dfrac{6+\sqrt{6}}{\sqrt{6}}$

$⇔ C=\dfrac{\sqrt{6}(\sqrt{6}+1)}{\sqrt{6}}$

$⇔ C=\sqrt{6}+1$

Chúc bạn học tốt !!!