Ta có: \(x=\frac{-1-\sqrt5}{2}\)

=>\(x^2=\frac{\left(-1-\sqrt5\right)^2}{2^2}=\frac{6+2\sqrt5}{4}=\frac{3+\sqrt5}{2}\)

=>\(x^4=\left(\frac{3+\sqrt5}{2}\right)^2=\frac{14+6\sqrt5}{4}=\frac{7+3\sqrt5}{2}\)

\(x^2\cdot x=\frac{\left(3+\sqrt5\right)\left(-\sqrt5-1\right)}{4}=\frac{-\left(3+\sqrt5\right)\left(\sqrt5+1\right)}{4}=\frac{-\left(3\sqrt5+3+5+\sqrt5\right)}{4}=-\sqrt5-2\)

=>\(x^3=-\sqrt5-2\)

\(x^5=x^3\cdot x^2=\left(-\sqrt5-2\right)\cdot\frac{3+\sqrt5}{2}\)

\(=-\frac12\left(2+\sqrt5\right)\left(3+\sqrt5\right)=-\frac12\left(6+2\sqrt5+3\sqrt5+5\right)=-\frac12\left(11+5\sqrt5\right)\)

\(4x^5+4x^4-5x^3+2x-2\)

\(=4\cdot\frac{-1}{2}\left(11+5\sqrt5\right)+4\cdot\frac{7+3\sqrt5}{2}-5\left(-\sqrt5-2\right)+2\cdot\frac{-1-\sqrt5}{2}\) -2

\(=-2\left(11+5\sqrt5\right)+2\left(7+3\sqrt5\right)+5\left(\sqrt5+2\right)-\sqrt5-1-2\)

\(=-22-10\sqrt5+14+6\sqrt5+5\sqrt5+10-\sqrt5-3=-1\)

Ta có: \(B=2\cdot\left(4x^5+4x^4-5x^3+2x-2\right)^{2021}+2022\)

=-2+2022

=2020

A(1/2^2022)=1/2^2022+1/2^4044+...+1/2^(2022^2021)

=>2^2022*A=1+1/2^2022+...+1/2^(2022^2020)

=>A*(2^2022-1)=1-1/2^(2022^2021)

=>\(A=\dfrac{2^{2022^{2021}}-1}{2^{2022}-1}\)

\(=\dfrac{tan\left(\dfrac{pi}{2}+x\right)\cdot sin\left(-x\right)\cdot cos\left(x-pi\right)}{cos\left(\dfrac{pi}{2}-x\right)\cdot sin\left(x+pi\right)}\)

\(=\dfrac{-cotx\cdot sin\left(-x\right)\cdot\left(-cosx\right)}{sinx\cdot-sinx}\)

\(=\dfrac{cotx\cdot sinx\left(-1\right)\cdot cosx}{-sinx\cdot sinx}=\dfrac{\dfrac{cosx}{sinx}\cdot cosx}{sinx}=\dfrac{cos^2x}{sin^2x}=cot^2x\)

A

đợi tý

Đã trả lời rồi còn độ tí đồ ngull