Đáp án:

$\begin{array}{l}
67)\\
a)\sqrt {36}  = \sqrt {{6^2}}  = 6\\
 - 3600; - 0,125\,\text{ko có căn bậc hai}\\
\sqrt {\frac{{36}}{{49}}}  = \sqrt {\frac{{{6^2}}}{{{7^2}}}}  = \frac{6}{7}\\
b) - 0,81\,\text{ko có căn bậc hai}\\
\sqrt {121}  = 11\\
\sqrt {{{\left( { - 0,81} \right)}^2}}  = \sqrt {0,{{81}^2}}  = 0,81\\
\sqrt {0,09}  = \sqrt {0,{3^2}}  = 0,3\\
c)\sqrt {\frac{{16}}{{81}}}  = \sqrt {\frac{{{4^2}}}{{{9^2}}}}  = \frac{4}{9}\\
\sqrt {\frac{{49}}{9}}  = \frac{7}{3}\\
 - \sqrt {{{\left( { - \frac{{52}}{{49}}} \right)}^2}}  =  - \frac{{52}}{{49}}\\
\sqrt {{{\left( {\frac{{52}}{{49}}} \right)}^2}}  = \frac{{52}}{{49}}\\
68)a)\sqrt 2  + \sqrt {11}  - \left( {\sqrt 3  + 5} \right)\\
 = \sqrt 2  - \sqrt 3  + \sqrt {11}  - 5\\
Do:\sqrt 2  < \sqrt 3 ;\sqrt {11}  < \sqrt {25}  = 5\\
 \Rightarrow \sqrt 2  - \sqrt 3  + \sqrt {11}  - 5 < 0\\
 \Rightarrow \sqrt 2  + \sqrt {11}  < \sqrt 3  + 5\\
b)\sqrt {21}  - \sqrt 5  - \left( {\sqrt {20}  - \sqrt 6 } \right)\\
 = \sqrt {21}  - \sqrt 5  - \sqrt {20}  + \sqrt 6 \\
 = \left( {\sqrt {21}  - \sqrt {20} } \right) + \left( {\sqrt 6  - \sqrt 5 } \right) > 0\\
 \Rightarrow \sqrt {21}  - \sqrt 5  > \sqrt {20}  - \sqrt 6 \\
69)a = \frac{{35}}{{49}} = \frac{5}{7}\\
b = \sqrt {\frac{{{5^2}}}{{{7^2}}}}  = \frac{5}{7}\\
c = \frac{{\sqrt {{5^2}}  + \sqrt {{{35}^2}} }}{{\sqrt {{7^2}}  + \sqrt {{{49}^2}} }} = \frac{{5 + 35}}{{7 + 49}} = \frac{{40}}{{56}} = \frac{5}{7}\\
d = \frac{{\sqrt {{5^2}}  - \sqrt {{{35}^2}} }}{{\sqrt {{7^2}}  - \sqrt {{{49}^2}} }} = \frac{{5 - 35}}{{7 - 49}} = \frac{{ - 30}}{{ - 42}} = \frac{5}{7}\\
 \Rightarrow a = b = c = d\\
70)a)\sqrt {{1^2}}  = 1\left( Đ \right)\\
b)\sqrt {{1^3} + {2^3}}  = \sqrt {1 + 8}  = \sqrt 9  = 3 = 1 + 2\\
 \Rightarrow \left( Đ \right)\\
c)\sqrt {{1^3} + {2^3} + {3^3}}  = \sqrt {1 + 8 + 27}  = \sqrt {36}  = 6\\
 = 1 + 2 + 3\\
 \Rightarrow \left( Đ \right)
\end{array}$

67)a)√36=√62=6−3600;−0,125ko có căn bậc hai√3649=√6272=67b)−0,81ko có căn bậc hai√121=11√(−0,81)2=√0,812=0,81√0,09=√0,32=0,3c)√1681=√4292=49√499=73−√(−5249)2=−5249√(5249)2=524968)a)√2+√11−(√3+5)=√2−√3+√11−5Do:√2<√3;√11<√25=5⇒√2−√3+√11−5<0⇒√2+√11<√3+5b)√21−√5−(√20−√6)=√21−√5−√20+√6=(√21−√20)+(√6−√5)>0⇒√21−√5>√20−√669)a=3549=57b=√5272=57c=√52+√352√72+√492=5+357+49=4056=57d=√52−√352√72−√492=5−357−49=−30−42=57⇒a=b=c=d70)a)√12=1(Đ)b)√13+23=√1+8=√9=3=1+2⇒(Đ)c)√13+23+33=√1+8+27=√36=6=1+2+3⇒(Đ)

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