A = 1 + 3 + 3² + 3³ + ... + 3⁶⁰

3A = 3 + 3² + 3³ + 3⁴ + ... + 3⁶¹

3A - A = (3 + 3² + 3³ + 3⁴ + ... + 3⁶¹) - (1 + 3 + 3² + 3³ + ... + 3⁶⁰)

2A = 3⁶¹ - 1

\(A=\frac{3^{61}-1}{2}\)

3A=3+3²+3³+3⁴+...+3⁶⁰+3⁶¹

3A - A=(3+3²+3³+3⁴+...+3⁶⁰+3⁶¹) - (1+3+3²+3³+....+3⁶⁰)

2A=3⁶¹-1

A  =\(\frac{3^{61}-1}{2}\)

\(3A=3+3^2+3^3+...+3^{61}\)

\(3A-A=3+3^2+3^3+...+3^{61}-1-3-3^2-...-3^{60}\)

\(2A=3^{61}-1\)

\(A=\frac{3^{61}-1}{2}\)

Chúc em học tốt^^

c)  2 3 + 5 3 - 60

= 6 + 15 - 2 15 = 6 - 15

          \(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=\)\(2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=\)\(6+\sqrt{15}-\sqrt{60}\)

\(=\)\(6-\sqrt{15}\)

\(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)

=\(6+\sqrt{15}-\sqrt{2^2.15}\)

=\(6+\sqrt{15}-2\sqrt{15}\)

=\(6-\sqrt{15}\)

k mk nha

Nhưng cho mk hỏi nhân kiểu nào ra 6 dc v ạ

$\frac{5a^2-33a+60}{a\left(a-1\right)\left(a-4\right)\left(a-5\right)}$

9√6 + 3√6 - √6 = 11√6

c) (√28 - 2√3 + √7)√7 + √84

= (√4.7 - 2√3 + √7)√7 + √4.21

= (2√7 - 2√3 + √7)√7 + 2√21

= (3√7 - 2√3)√7 + 2√21

= 3.7 - 2√21 + 2√21 = 21

\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(4\sqrt{3}+\sqrt{5}\right)^2}\)

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

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

\(\sqrt{\left(4+\sqrt{3}\right)\sqrt{19-8\sqrt{3}}+3}=\sqrt{\left(4+\sqrt{3}\right)\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

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

a) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{60}}\)

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

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

b) Ta có: \(\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{19-8\sqrt{3}+3}}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\left(4-\sqrt{3}\right)+3}\)

=4

a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

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

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

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

Ta có √[ 5 + 2 + 3 + 2√(2×3) + 2√(2×5) + 2√(3×5)] = √[(√2 + √3 + √5)²] = √2 + √3 + √5

Ta có √[ 5 + 2 + 3 + 2√(2×3) + 2√(2×5) + 2√(3×5)] = √[(√2 + √3 + √5)²] = √2 + √3 + √5

3C

4B

5A

3. C

4. B

5. A

3. C

4. B

5. A