3.(x+4)= 23x+2
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28 tháng 6 2021
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)
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3. \(Pt\Leftrightarrow\left(sinx+sin3x\right)+\left(sin2x+sin4x\right)=0\)
\(\Leftrightarrow2.sin2x.cosx+2.sin3x.cosx=0\)
\(\Leftrightarrow2cosx\left(sin2x+sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin3x=-sin2x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\sin3x=sin\left(\pi+2x\right)\end{matrix}\right.\)(\(k\in Z\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pi+k2\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\)(\(k\in Z\))\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\dfrac{k2\pi}{5}\end{matrix}\right.\) (\(k\in Z\))
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4. Pt\(\Leftrightarrow\dfrac{1-cos2x}{2}+\dfrac{1-cos4x}{2}=\dfrac{1-cos6x}{2}\)
\(\Leftrightarrow cos2x+cos4x=1+cos6x\)
\(\Leftrightarrow2cos3x.cosx=2cos^23x\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=0\\cosx=cos3x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=-k\pi\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x=\dfrac{k\pi}{2}\end{matrix}\right.\)\(\left(k\in Z\right)\)
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16 tháng 4 2017
\(\Rightarrow\dfrac{14}{3}x-\dfrac{7}{4}\)=\(\dfrac{1}{23x}\)
\(\dfrac{14}{3}x-\dfrac{1}{23x}\)=\(\dfrac{7}{4}\)
23 tháng 9 2021
d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
15 tháng 7 2019
\(a)x^4-2x^3-3x^2+4x+4=(x^4-x^3-2x^2)-\left(x^3-x^2-2x\right)-\left(2x^2-2x-4\right)\)
\(=\left(x^2-x-2\right)\left(x^2-x-2\right)=\left(x^2-x-2\right)^2\)
\(b)x^4+2x^3-23x^2-24x+144=\left(x^4+x^3-12x^2\right)+\left(x^3+x^2-12x\right)-\left(12x^2+12x-144\right)\)
\(=\left(x^2+x-12\right)\left(x^2+x-12\right)=\left(x^2+x-12\right)^2\)
2 tháng 4 2023
`G(x)+H(x)=(21x^2+1+17x)+(-2+6x^3-12x^2-8)`
`=21x^2+1+17x-2+6x^3-12x^2-8`
`= 6x^3+(21x^2-12x^2)+17x+(1-2-8)`
`= 6x^3+9x^2+17x-9`
`G(x)-H(x)=(21x^2+1+17x)-(-2+6x^3-12x^2-8)`
`= 21x^2+1+17x+2-6x^3+12x^2+8`
`= -6x^3+(21x^2+12x^2)+17x+(1+2+8)`
`= -6x^3+33x^2+17x+11`
`----`
`M(x)+N(x)=(7x^5 + 1 + 17x^4 - 2)+(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2+6x^4 - 12x^2 - 23x^4 + x`
`= 7x^5+(17x^4+6x^4-23x^4)-12x^2+x+(1-2)`
`= 7x^5-12x^2+x-1`
`M(x)-N(x)=(7x^5 + 1 + 17x^4 - 2)-(6x^4 - 12x^2 - 23x^4 + x)`
`= 7x^5 + 1 + 17x^4 - 2-6x^4 + 12x^2 + 23x^4 - x`
`= 7x^5+(17x^4-6x^4+23x^4)+12x^2-x+(1-2)`
`= 7x^5+34x^4+12x^2-x-1`
2 tháng 4 2023
Mình đã trl rồi nha!
(https://hoc24.vn/cau-hoi/tinh-tong-avf-hieu-cac-da-thuc-saugx-21x2-1-17x-va-hx-2-6x3-12x2-8mx-7x5-1-17x4-2-va-nx-6x4-12x2-23x4-x.7858748287383)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2020
\(\Leftrightarrow x^4-4x^3+4x^2-4x^3+16x^2-16x+3x^2-12x+12\le0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)-4x\left(x^2-4x+4\right)+3\left(x^2-4x+4\right)\le0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x-2\right)^2\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2-4x+3\le0\end{matrix}\right.\) \(\Rightarrow1\le x\le3\)
3(x + 4) = 23x + 2
<=> 3x + 12 = 23x + 2
<=> 3x - 23x = 2 - 12
<=> -20x = -10
<=> x = 0,5
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