\(B=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\)

\(2B=2\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{200}{2^{200}}\right)\)

\(2B=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{200}{2^{199}}\)

\(2B-B=\left(2+\frac{3}{2^2}+...+\frac{200}{2^{199}}\right)-\left(1+\frac{3}{2^3}+...+\frac{200}{2^{200}}\right)\)

.... đặt A=... giiả tiếp

a) Ta có: A = 1 + 3 + + ... +

⇒ 3A = 3 + + ... +

⇒ 3A - A = (3 + + ... + 1 + 3 + + ... +

⇒ 2A = 3²⁰¹ - 1

⇒ A = 2A : A = \(\dfrac{3^{201}-1}{2}\)

Vậy A = \(\dfrac{3^{201}-1}{3}\)

a: \(3A=3+3^2+3^3+...+3^{201}\)

=>\(2A=3^{201}-1\)

hay \(A=\dfrac{3^{201}-1}{2}\)

b: \(B=4+4^2+4^3+...+4^{200}\)

=>\(4B=4^2+4^3+4^4+...+4^{201}\)

=>\(3B=4^{201}-4\)

hay \(B=\dfrac{4^{201}-4}{3}\)

Q=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\)\(\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3-2}}\)=\(\frac{1}{\sqrt{2}-\sqrt{3}}\).\(\left(\sqrt{3}-\sqrt{2}\right)^{ }\)=-1

B= -1/3+1/3^2-1/3^3+…+1/3^100-1/3^101

3B= -1+1/3-1/3^2+…+1/3^99-1/3^100

3B+B=4B=-1-1/3^101

=>B=(-1-1/3^101)/4

Vậy B=(-1-1/3^101)/4

Ta có: \(E=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)

\(=\frac{1}{\sqrt{2}-\sqrt{3}}\cdot\sqrt{\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}}\)

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

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

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

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

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

\(A=x^2-7x^3+7\)Thay x = 1 ta được 

\(A=1-7+7=1\)

Thay x=1 vào A ta có:
\(A=2x^2-4x^3+7-x^2-3x^3\\ =x^2-7x^3+7\\ =1^2-7.1^3+7\\ =1-7+7\\ =1\)

thay x-1 vào A ta đc :

\(A=2.1^2-4.1^3+7-1^2-3.1^3\)

=1

HT

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

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

ok???