√3 cos2x + 2sinx.cosx = -1
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cos 2 x + 2 sin x . cos x + 5 sin 2 x = 2
Rõ ràng cosx = 0 không thỏa mãn phương trình. Với cosx ≠ 0, chia hai vế cho cos2x ta được:
1 + 2 tan x + 5 tan 2 x = 2 ( 1 + tan 2 x ) ⇔ 3 tan 2 x + 2 tan x - 1 = 0
\(3\cdot(-2)^{5x+8}+5^2=1\\\Rightarrow 3\cdot(-2)^{5x+8}+25=1\\\Rightarrow 3\cdot(-2)^{5x+8}=1-25\\\Rightarrow 3\cdot(-2)^{5x+8}=-24\\\Rightarrow (-2)^{5x+8}=-24:3\\\Rightarrow (-2)^{5x+8}=-8\\\Rightarrow (-2)^{5x+8}=(-2)^3\\\Rightarrow 5x+8=3\\\Rightarrow 5x=3-8\\\Rightarrow 5x=-5\\\Rightarrow x=-5:5\\\Rightarrow x=-1\)
(x-1)^3-x(x-2)^2+1
= x^3-3x^2+3x-1-x(x^2-4x+4)+1
= x^3-3x^2+3x-1- x^3+4x^2-4x+1
= x^2-x
= x(x-1)
HỌC TỐT!
@Zịt_siu_lừi
\(=x^3-3x^2+3x-1-x\left(x^2-4x+4\right)+1\)
\(=x^3-3x^2+3x-1-x^3+4x^2-4x+1\)
\(=x^2-x\)
\(\Leftrightarrow\left|3x-2\right|>x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x-2>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\3x-2>x^2+2x+1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2+2x+1-3x+2< 0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>-1\\x^2-x+3< 0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
c/
\(\Leftrightarrow\sqrt{2}sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{\sqrt{2}}\)
\(\Leftrightarrow sin\left(3x-\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}3x-\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\3x-\frac{\pi}{4}=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7\pi}{36}+\frac{k2\pi}{3}\\x=\frac{11\pi}{36}+\frac{k2\pi}{3}\end{matrix}\right.\)
d/
\(\Leftrightarrow2sinx.cosx+1-2sin^2x=1\)
\(\Leftrightarrow2sinx\left(cosx-sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=cosx\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
a/
\(\Leftrightarrow\frac{\sqrt{3}}{2}sin5x-\frac{1}{2}cos5x=-1\)
\(\Leftrightarrow sin\left(5x-\frac{\pi}{6}\right)=-1\)
\(\Leftrightarrow5x-\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=-\frac{\pi}{15}+\frac{k2\pi}{5}\)
b/
\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{3}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{3}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
`\sqrt{3}cos 2x+2sin x cos x=-1`
`<=>\sqrt{3}cos 2x+sin 2x=-1`
`<=>cos(2x-\pi/6)=-1/\sqrt{2}`
`<=>[(2x-\pi/6=[3\pi]/4+k2\pi),(2x-\pi/6=[-3\pi]/4+k2\pi):}`
`<=>[(x=[11\pi]/24+k\pi),(x=[-7\pi]/24+k\pi):}` `(k in ZZ)`