Question

cho tanx=2 tính Q=\(\dfrac{\sin^3x}{2\sin x+\cos^3x}\)

Answer 1

tan x=2

=>\(\dfrac{sinx}{cosx}=2\)

=>sin x và cosx cùng dấu và \(sinx=2\cdot cosx\)

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+4=5\)

=>\(cos^2x=\dfrac{1}{5}\)

=>\(\left[{}\begin{matrix}cosx=\dfrac{1}{\sqrt{5}}\\cosx=-\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)

TH1: \(cosx=\dfrac{1}{\sqrt{5}}\)

=>\(sinx=\sqrt{1-cos^2x}=\dfrac{2}{\sqrt{5}}\)

TH2: cosx=-1/căn 5

=>\(sinx=-\sqrt{1-cos^2x}=-\dfrac{2}{\sqrt{5}}\)

\(Q=\dfrac{sin^3x}{2sinx+cos^3x}\)

\(=\dfrac{\left(2\cdot cosx\right)^3}{2\cdot2cosx+cos^3x}\)

\(=\dfrac{8\cdot cos^3x}{4cosx+cos^3x}=\dfrac{8cos^2x}{4+cos^2x}\)

\(=\dfrac{8\cdot\dfrac{1}{5}}{4+\dfrac{1}{5}}=\dfrac{8}{5}:\dfrac{21}{5}=\dfrac{8}{21}\)