3cos^x - 5cosx + 2 = 0 3tan^x - 2căn3tanx + 3 = 0 Giải giúp mk 2 câu này với
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1.
\(\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)+5cosx+3=0\)
\(\Leftrightarrow2cos^2x-1+5cosx+3=0\)
\(\Leftrightarrow2cos^2x+5cosx+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\frac{1}{2}\\cosx=-2\left(ktm\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
\(3\left(1-sin^2x\right)+\left(1-sin^2x\right)sinx=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx\right)+\left(1+sinx\right)\left(sinx-sin^2x\right)=8\left(1+sinx\right)\)
\(\Leftrightarrow\left(1+sinx\right)\left(3-3sinx+sinx-sin^2x-8\right)=0\)
\(\Leftrightarrow\left(1+sinx\right)\left(-sin^2x-2sinx-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=-1\\-sin^2x-2sinx-5=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
\(A=\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\)
\(=\left(\dfrac{3}{\sqrt{x}-2}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{\left(\sqrt{x}+3\right)^2}\)
\(=\dfrac{3\sqrt{x}+9-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+13}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
Vậy...
Pt: \(\Rightarrow-3\left(cos^2x-sin^2x\right)-\sqrt{3}sin2x=0\)
\(\Rightarrow-3cos2x-\sqrt{3}sin2x=0\)
\(\Rightarrow sin2x+\sqrt{3}cos2x=0\)
\(\Rightarrow2sin\left(2x+\dfrac{\pi}{3}\right)=0\) \(\Rightarrow sin\left(2x+\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow2x+\dfrac{\pi}{3}=k\pi\left(k\in Z\right)\)
\(\Rightarrow x=-\dfrac{\pi}{6}+k\dfrac{\pi}{2}\)
|x+2|<3
\(\Rightarrow-3\le x+2\le3\)3
\(\Rightarrow-1\le x\le1\)
\(\Rightarrow x=-1;0;1\)
\(a,3x-6=5x+2\)
\(3x-5x=2+6\)
\(-2x=8\)
\(x=-4\)
\(b,2\times\left(x-3\right)-3\times\left(x+7\right)=14\)
\(2x-6-3x-21=14\)
\(-x-27=14\)
\(x=-27-14\)
\(x=-41\)
a) 3x - 6 = 5x + 2
3x - 6 - 2 = 5x
3x - 8 = 5x
3x - 5x = 8
-2x = 8
x = -4
b) 2(x - 3) - 3(x + 7) = 14
2x - 6 - 3x - 21 = 14
(-x) - 27 = 14
(-x) = 41
x = -41
c) 3x2 - x - 2 = 0
x.(3x - 1) = 2
x.(3x - 1) = 2 = 1.2 = 2.1 = (-1).(-2) = (-2).(-1)
Xét 4 trường hợp ,ta có :
\(\left(1\right)\hept{\begin{cases}x=1\\3x-1=2\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=1\end{cases}}}\)(nhận)
\(\left(2\right)\hept{\begin{cases}x=2\\3x-1=1\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{2}{3}\end{cases}}}\)(loại)
\(\left(3\right)\hept{\begin{cases}x=-1\\3x-1=-2\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-\frac{1}{3}\end{cases}}}\)(loại)
\(\left(4\right)\hept{\begin{cases}x=-2\\3x-1=-1\end{cases}\Rightarrow\hept{\begin{cases}x=-2\\x=0\end{cases}}}\)(loại)