\(\dfrac{4}{\sqrt{7}+\sqrt{3}}+\dfrac{4}{\sqrt{7}-\sqrt{3}}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}\\ =\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}+\dfrac{4\left(\sqrt{7}+\sqrt{3}\right)}{4}\\ =\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\\ =2\sqrt{7}\)

@seven

\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)

=2*căn 7

C = 5³ + 5⁵ + ... + 5¹⁰¹

⇒ 25C = 5⁵ + 5⁷ + ... + 5¹⁰¹

⇒ 24C = 25C - C

= (5⁵ + 5⁷ + ... + 5¹⁰³) - (5³ + 5⁵ + ... + 5¹⁰¹)

= 5¹⁰³ - 5⁵

⇒ C = (5¹⁰³ - 5⁵)/24

--------

D = 1 + 3² + 3⁴ + ... + 3¹⁰⁰

⇒ 9D = 3² + 3⁴ + 3⁶ + ... + 3¹⁰²

⇒ 8D = 9D - D

= (3² + 3⁴ + 3⁶ + ... + 3¹⁰²) - (1 + 3² + 3⁴ + ... + 3¹⁰⁰)

= 3¹⁰² - 1

⇒ D = (3¹⁰² - 1)/8

tôi gửi lại, thông cảm

\(D=\sqrt{3+2\sqrt{3}+1}+\sqrt{4-2\cdot2\sqrt{3}+3}-\sqrt{9+2\cdot3\cdot\sqrt{3}+3}\)

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

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

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

giúp mình với mình đang rất cần các bạn giúp đỡ.

\(C=\dfrac{14}{33}\cdot a\cdot x^4y^5+\dfrac{5}{2}ab\cdot x^3y^4z+ax\cdot x^6y^3\)

\(=\dfrac{14}{33}a\cdot x^4y^5+\dfrac{5}{2}ab\cdot x^3y^4z+a\cdot x^6y^3\)

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)

Ta có: ( Sửa đề )

\(A=4+4^2+4^3+...+4^{2021}+4^{2022}\)

\(A=\left(4+4^2\right)+\left(4^3+4^4\right)+...+\left(4^{2021}+4^{2022}\right)\)

\(A=20+4^2.\left(4+4^2\right)+...+4^{2020}.\left(4+4^2\right)\)

\(A=20+4^2.20+...+4^{2020}.20\)

\(A=20.\left(1+4^2+...+4^{2020}\right)\)

Vì \(20⋮20\) nên \(20.\left(1+4^2+...+4^{2020}\right)\)

Vậy \(A⋮20\)

\(#WendyDang\)