\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)
= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)=\(\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\)
(2cos2x+1)(2cos6x+1)(2cos18x+1)=1
\(\frac{1}{2^2-1}\)+\(\frac{1}{3^2-1}\)+....+\(\frac{1}{n^2-1}\) bẳng ?
giải pt : 1, \(\dfrac{\left(1-cosx\right)^2+\left(1+sinx\right)^2}{4\left(1-sinx\right)}-tan^2sinx=\dfrac{1}{2}\left(1+sinx\right)+tan^2x\)
Giải các phương trình sau:
1. tan2x+3= (1+√2 sin x)(tan x+ √2 cos x)
2. (1- cos x. cos2x )/ sin2x - 1/ cos x= 4 sin2x - sin x-1
3. sin3x + 2 cos3x+ cos2x - 2sin2x - 2sinx-1=0
\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+...+\(\frac{1}{2^n}\)=\(\frac{2^n-1}{2^n}\)
\(\left(2cos4x+2cosx+1\right)\left(2cos20x+2cos10x+1\right)=1\)
cho hàm số y=cosx+msinx+1/cosx+2 (1). xác định m để GTLN của hàm số (1) trên tập R bằng 1
U1 = 2
Un+1 = \(\frac{1}{3}\)(Un+1) Tìm U4