\(1,\)

\(a,\sqrt{6-2\sqrt{5}}=\sqrt{\sqrt{5^2}-2.\sqrt{5}.1+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

\(b,\sqrt{8+2\sqrt{7}}=\sqrt{\sqrt{7^2}+2.\sqrt{7}.1+1}=\sqrt{\left(\sqrt{7}+1\right)^2}=\left|\sqrt{7}+1\right|=\sqrt{7}+1\)

\(2,\)

\(a,\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{10}\)

\(=\left|\sqrt{10}-3\right|-\sqrt{10}\)

\(=\sqrt{10}-\sqrt{10}-3\)

\(=-3\)

\(b,\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}\)

\(=\left|5+\sqrt{7}\right|-\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=5+\sqrt{7}-\left|\sqrt{7}-1\right|\)

\(=5+\sqrt{7}-\sqrt{7}+1\)

\(=6\)

Em cố gắng gõ latex nha

a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{2}{3}\)

b: \(=\dfrac{11^9\cdot3^{18}}{3^{18}\cdot11^{11}}=\dfrac{1}{121}\)

a: \(=\dfrac{20\left(1-12\right)}{30\left(-1-10\right)}=\dfrac{20}{30}=\dfrac{2}{3}\)

b: \(=\dfrac{11^9\cdot3^{18}}{3^{10}\cdot11^{11}\cdot3^8}=\dfrac{1}{121}\)

a: =8/4-1/4=7/4

b: =3/9=1/3

c: =5/7-4/7=1/7

a,\(\dfrac{7}{4}\)

b,\(\dfrac{1}{3}\)

c,\(\dfrac{1}{7}\)

a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)

\(=8\cdot5^2+5-1\)

\(=200+4=204\)

b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)

\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)

\(=27-1+125-1\)

=150

a. (8,5⁷ - 5⁵ + 5⁶) : 5⁵

= (8,5⁷ : 5⁵) - (5⁵ : 5⁵) + (5⁶ : 5⁵)

= 1,7² - 1 + 5

= 2,89 - 1 + 5

= 6,89

b. (9³⁰ - 27¹⁹) : 3⁵⁷ + (125⁹ - 25¹²) : 5²⁴

= (9³⁰ : 3⁵⁷) - (27¹⁹ : 3⁵⁷) + (125⁹ : 5²⁴) - (25¹² : 5²⁴)

= 3³ - 1 + 125 - 1

= 27 - 1 + 125 - 1

= 150

c. (10¹² + 5¹¹ . 2⁹ - 5¹³ - 2⁸) : 4 . 5⁵ . 10⁶

= (10¹² + 2,5 , 10¹⁰ - 5¹³ - 2⁸) : 1,25 . 10¹⁰

= (10¹² : 1,25 . 10¹⁰) + (2,5 . 10¹⁰ : 1,25 . 10¹⁰) - (5¹³ : 1,25 . 10¹⁰) - (2⁸ : 1,25 . 10¹⁰)

= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)

= 81,90234373 \(\approx\) 82

1:

a: Vì \(\dfrac{-4}{3}=\dfrac{-4\cdot3}{3\cdot3}=\dfrac{-12}{9}=\dfrac{12}{9}\\ \Rightarrow\dfrac{-4}{3}=\dfrac{12}{9}\)

b: Vì : \(-2\cdot3=-6\\ -6\cdot8=-48\)

nên 2 p/s ko bằng nhau 

\(a,=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\\ b,=\left|1-\sqrt{2}\right|+\sqrt{5}=\sqrt{2}-1+\sqrt{5}\)

a)\(\sqrt{12}+3\sqrt{27}-\sqrt{300}=2\sqrt{3}+9\sqrt{3}-10\sqrt{3}=\sqrt{3}\)

b) \(\sqrt{\left(1-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}+3\right)^2}=\left|1-\sqrt{2}\right|-\left|\sqrt{2}+3\right|=\sqrt{2}-1-\sqrt{2}-3=-4\)

a) \(A=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}-\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}-1\right|-\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+-\sqrt{3}-1=-2\)

b) \(B=\sqrt{11-6\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{3^2-2.3.\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}\)

\(=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=\left|3-\sqrt{2}\right|-\left|\sqrt{2}-1\right|\)

\(=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)

c) \(C=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{5}.\sqrt{2}+\left(\sqrt{2}\right)^2}\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{5}+\sqrt{3}\right)\left|\sqrt{5}-\sqrt{2}\right|\)

\(=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{2}\right)=5-\sqrt{10}+\sqrt{15}-\sqrt{6}\)

Sao ko bảo chị giải đi, kkkk

Haizzzzz

Câu a)

Câu b) Đề bài sai.