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Lời giải:
Mệnh đề sai, do với $x=0\in\mathbb{R}$ thì $x^2=0$
Mệnh đề phủ định:
$\overline{A}: \exists x\in\mathbb{R}, x^2\leq 0$
a: ĐKXĐ: x>=9/4
\(\sqrt{4x-9}=2x-5\)
=>\(\begin{cases}\left(2x-5\right)^2=4x-9\\ 2x-5\ge0\end{cases}\Rightarrow\begin{cases}4x^2-20x+25-4x+9=0\\ x\ge\frac52\end{cases}\)
=>\(\begin{cases}4x^2-24x+34=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}2x^2-12x+17=0\\ x\ge\frac52\end{cases}\)
=>\(\begin{cases}x^2-6x+\frac{17}{2}=0\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x^2-6x+9-\frac12=0\\ x\ge\frac52\end{cases}\)
=>\(\begin{cases}\left(x-3\right)^2=\frac12\\ x\ge\frac52\end{cases}\Rightarrow\begin{cases}x-3\in\left\lbrace\frac{\sqrt2}{2};-\frac{\sqrt2}{2}\right\rbrace\\ x\ge\frac52\end{cases}\)
=>\(x=3+\frac{\sqrt2}{2}=\frac{6+\sqrt2}{2}\)
b: ĐKXĐ: \(x^2-7x+10\ge0\)
=>(x-5)(x-2)>=0
=>x>=5 hoặc x<=2
\(\sqrt{x^2-7x+10}=3x-1\)
=>\(\begin{cases}3x-1\ge0\\ \left(3x-1\right)^2=x^2-7x+10\end{cases}\Rightarrow\begin{cases}9x^2-6x+1-x^2+7x-10=0\\ x\ge\frac13\end{cases}\)
=>\(\begin{cases}8x^2+x-9=0\\ x\ge\frac13\end{cases}\Rightarrow\begin{cases}8x^2+9x-8x-9=0\\ x\ge\frac13\end{cases}\)
=>\(\begin{cases}\left(x+1\right)\left(8x-9\right)=0\\ x\ge\frac13\end{cases}\Rightarrow x=\frac98\) (nhận)
d: |3x-1|=x+3
=>\(\begin{cases}x+3\ge0\\ \left(3x-1\right)^2=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-3\\ \left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\end{cases}\)
=>\(\begin{cases}x\ge-3\\ \left(2x-4\right)\left(4x+2\right\rbrace\end{cases}\Rightarrow x\in\left\lbrace2;-\frac12\right\rbrace\)
e: |x+2|=|6-3x|
=>|3x-6|=|x+2|
=>3x-6=x+2 hoặc 3x-6=-x-2
=>2x=8 hoặc 4x=4
=>x=4 hoặc x=1
\(A=sin\left(32+28\right)=sin60=\frac{\sqrt{3}}{2}\)
\(B=cos\left(26+4\right)=cos30=\frac{\sqrt{3}}{2}\)
\(\dfrac{sina+sin5a+sin3a}{cosa+cos5a+cos3a}=\dfrac{2sin3a.cos2a+sin3a}{2cos3a.cos2a+cos3a}=\dfrac{sin3a\left(2cos2a+1\right)}{cos3a\left(2cos2a+1\right)}=\dfrac{sin3a}{cos3a}=tan3a\)
\(\dfrac{1+sin4a-cos4a}{1+sin4a+cos4a}=\dfrac{1+2sin2a.cos2a-\left(1-2sin^22a\right)}{1+2sin2a.cos2a+2cos^22a-1}=\dfrac{2sin2a\left(sin2a+cos2a\right)}{2cos2a\left(sin2a+cos2a\right)}=\dfrac{sin2a}{cos2a}=tan2a\)
\(96\sqrt{3}sin\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{48}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=48\sqrt{3}sin\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{24}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=24\sqrt{3}sin\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{12}\right)cos\left(\dfrac{\pi}{6}\right)=12\sqrt{3}sin\left(\dfrac{\pi}{6}\right)cos\left(\dfrac{\pi}{6}\right)\)
\(=6\sqrt{3}sin\left(\dfrac{\pi}{3}\right)=6\sqrt{3}.\dfrac{\sqrt{3}}{2}=9\)
\(A+B+C=\pi\Rightarrow A+B=\pi-C\Rightarrow tan\left(A+B\right)=tan\left(\pi-C\right)\)
\(\Rightarrow\dfrac{tanA+tanB}{1-tanA.tanB}=-tanC\Rightarrow tanA+tanB=-tanC+tanA.tanB.tanC\)
\(\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC\)
\(A=sinx+cosx=\sqrt{2}sin\left(x+\frac{\pi}{4}\right)\)
Mà \(-1\le sin\left(x+\frac{\pi}{4}\right)\le1\Rightarrow-\sqrt{2}\le sinx+cosx\le\sqrt{2}\)
\(A_{max}=\sqrt{2}\) khi \(sin\left(x+\frac{\pi}{4}\right)=1\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\Rightarrow x=\frac{\pi}{4}+k2\pi\)
\(A_{min}=-\sqrt{2}\) khi \(x+\frac{\pi}{4}=-\frac{\pi}{2}+k2\pi\Rightarrow x=-\frac{3\pi}{4}+k2\pi\)
2 câu sau y hệt câu đầu:
\(B=sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\Rightarrow-\sqrt{2}\le B\le\sqrt{2}\)
\(C=sin4x+cos4x=\sqrt{2}sin\left(4x+\frac{\pi}{4}\right)\Rightarrow-\sqrt{2}\le C\le\sqrt{2}\)