=1                 

1, 3^4

2, 5^405

3, -20/3

4, 4

                                                                             KHÔNG BÍT ĐÚNG HAY KO =>HIHIHIHI

Lời giải:
Biểu thức \(=\frac{(5+2\sqrt{6})(25+24-2\sqrt{25.24})\sqrt{3+2-2\sqrt{3.2}}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(\sqrt{25}-\sqrt{24})^2.\sqrt{(\sqrt{3}-\sqrt{2})^2}}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})^2(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5+2\sqrt{6})(5-2\sqrt{6})(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}\)

\(=\frac{(5-2\sqrt{6})(\sqrt{3}-\sqrt{2})}{9\sqrt{3}-11\sqrt{2}}=\frac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}=1\)

Câu 1 xem kỉ đề

\(B,\frac{49^6.5-7^{11}}{\left(-7\right)^{10}.5-2.49^5}=\frac{7^{12}.5-7^{11}}{7^{10}.5-2.7^{10}}=\frac{7^{11}.\left(7.5-1\right)}{7^{10}.\left(5-2\right)}=\frac{7.34}{3}=\frac{238}{3}\)

a) A=2¹².3⁵-\(\frac{2^{12}.3^6}{2^{12}}\)+9³+8⁴.3⁵

=2¹².3⁵-3⁶+3⁶+2¹².3⁵

=2¹³.3⁵

^b)B=49⁶.5-5.\(\frac{7^{11}}{\left(-7\right)^{10}}-2.49^5\)

=49⁶.5-7.5-2.49⁵

⁼7¹².5-7.5-2.7¹⁰

\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)

\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)

a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)

\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)

\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)

\(=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\)

\(=-6\sqrt{3}\)

Câu a:

\(\frac{2^{12}.3^5-4^6.9^2}{12^6+8^4.3^5}\)

= \(\frac{2^{12}.3^5-2^{12.3^6}}{2^{12}.3^6+2^{12}.3^5}\)

= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}\)

= \(\frac{2}{3.4}\)

= \(\frac16\)

Câu b:

\(\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

= \(\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

= \(\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+8\right)}\)

= \(\frac{5.\left(-6\right)}{9}\)

=- 10/3

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$

$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$