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

giup mk di

nhanh len cac ban oi

Viết lại đề cho bạn nè
  \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

= \(\sqrt{9-6\sqrt{3}+3}-\sqrt{12-12\sqrt{3}+9}\)

= \(\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}\right)^2-2.3.\sqrt{12}+3^2}\)

= \(\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)

= |\(3-\sqrt{3}\)| - |\(\sqrt{12}-3\)|

= \(3-\sqrt{3}-\sqrt{12}+3\)

= \(6-\sqrt{3}-2\sqrt{3}\)

= \(6-3\sqrt{3}\)

\(\frac{2}{9}.x=\frac{6}{11}\)

\(\Leftrightarrow x=\frac{6}{11}:\frac{2}{9}\)

\(\Rightarrow x=\frac{54}{22}\)

2/9 * x = 6/11

x= 6 / 11 : 2/9

x= 6/11 * 9/2

x= 54/22

x=6/11:2/9=27/11

Đặt A = 1 + 7 + 7²+...+7¹⁰¹

=> 7A = 7 + 7²+7³+...+7¹⁰²

=> 7A-A= 7¹⁰²-1

6A = 7¹⁰²-1

\(\Rightarrow A=\frac{7^{102}-1}{6}\)

Đặt \(A=1+7^2+7^3+7^4+...+7^{99}\)

\(\Rightarrow7A=7\left(1+7^2+7^3+7^4+...+7^{99}\right)\)

\(\Rightarrow7A=7+7^3+7^4+7^5+...+7^{100}\)

\(\Rightarrow7A-A=\left(7+7^3+7^4+7^5+...+7^{100}\right)-\left(1+7^2+7^3+7^4+...+7^{99}\right)\)

\(\Rightarrow6A=7^{100}-1\)

\(\Rightarrow A=\frac{7^{100}-1}{6}\)

đặt S = 1 + 7² + 7³ + 7⁴ + .... + 7⁹⁹

=> 7S = 7 + 7³ + 7⁴ + 7⁵ + .... 7¹⁰⁰

=> 7S - S = (7 + 7³ + 7⁴ + 7⁵ + .... + 7¹⁰⁰) - (1 + 7² + 7³ + 7⁴ + .... + 7⁹⁹)

=> 6S = (7 + 7¹⁰⁰) - (1 + 7²) 

=> 6S = (7 - 1) + (7¹⁰⁰ - 7²)

=> 6S = 6 + 7¹⁰⁰ - 7²

=> S = \(\frac{6+7^{100}-7^2}{6}\)

rốt cuộc là ai đúng

\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\frac{\sqrt{1}}{\sqrt{8a^2}}=\frac{1}{\sqrt{2}\sqrt{4}\sqrt{a}}\)

=\(\frac{1}{2\sqrt{2}a}\)