Giải pt: 9x^4+23x-12=0
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a) \(\text{Δ}=8^2-4.3.4=16\)
\(\left[{}\begin{matrix}x=\dfrac{-8+4}{2.3}=-\dfrac{2}{3}\\x=\dfrac{-8-4}{2.3}=-2\end{matrix}\right.\)

2x(3x-4)-9x+12=0
<=>2x(3x-4)-3(3x-4)
<=>(3x-4)(2x-3)
<=>3x-4=0 hoặc 2x-3=0
- 3x-4=0<=>3x=4<=>x=4/3
- 2x-3=0<=>2x=3<=>x=1,5
Vậy S={4/3;1,5}

\(\Leftrightarrow3.\left(1-cos6x\right)+2.cos^26x-1=4\)
\(\Leftrightarrow2.cos^26x-3.cos6x-2=0\)
\(\Leftrightarrow\left(cos6x-2\right)\left(2.cos6x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos6x=2\left(vn\right)\\cos6x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}6x=\dfrac{2\pi}{3}+k2\pi\\6x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\) (\(k\in Z\)) \(\Leftrightarrow\)\(\left[{}\begin{matrix}x\ne\dfrac{\pi}{9}+\dfrac{k\pi}{3}\\x\ne\dfrac{-\pi}{9}+\dfrac{k\pi}{3}\end{matrix}\right.\) (\(k\in Z\))
Vậy...


1.Pt \(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=sin\left(x+\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\left(\dfrac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\\2x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
\(\Rightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\)\(\left(k\in Z\right)\)
2.\(sin^22x+cos^23x=1\)
\(\Leftrightarrow\dfrac{1-cos4x}{2}+\dfrac{1+cos6x}{2}=1\)
\(\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{5}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow x=\dfrac{k\pi}{5}\)\(\left(k\in Z\right)\) (Gộp nghiệm)
Vậy...
3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
Vậy...
4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
Vậy...

\(9x^4-4=0\\ \Leftrightarrow\left(3x^2\right)^2-2^2=0\\ \Leftrightarrow\left(3x-2\right)\left(3x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy...
IQ 4 tỷ
(3x2)2 - 22=3x2 - 2 chớ ta
\(\left\{{}\begin{matrix}3x^2-2=0\\3x^2+2=0\end{matrix}\right.\left\{{}\begin{matrix}x^2=\frac{2}{3}\\x^2=-\frac{2}{3}\left(loại\right)\end{matrix}\right.\left\{{}\begin{matrix}x=\sqrt{\frac{2}{3}}\\\end{matrix}\right.\)

a) X^3-x^2-21x+45=0
x^3-3x^2+2x^2-6x-15x+45=0
x^2(x-3)+2x(x-3)-15(x-3)=0
(x-3)(x^2+2x-15)=0
(x-3)(x^2-3x+5x-15)=0
(x-3)[x(x-3)+5(x-3)]=0
(x-3)^2(x+5)=0
<=> x=3 hoặc x=-5
Câu 2 đề ko rõ lắm bn sửa lại đề để mk giải hộ nha
Bích Ngọc bạn xem lời giải dưới đây nhé :
X^3-x^2-21x+45=0\(\Leftrightarrow\)(x+5)(x^2-6x+9)=0
\(\Leftrightarrow\)(x+5)(x-3)^2=0
Rồi đó tới đây bạn tự tìm x nhé!

tương tự như phần vừa nãy nha bạn tự giải được kết quả x=-1 và x=4 là đúng
uh