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`y'=[3(x+1)-3x-2]/[(x+1)^2]=1/[(x+1)^2]`
Gọi `M(x_0; y_0)-` tiếp điểm
Mà `y_0=[3x_0+2]/[x_0+1] in T T`
`=>y-[3x_0+2]/[x_0+1]=1/[(x_0+1)^2](x-x_0)`
`@` Gọi `T T nn Ox =A`
`=>-[3x_0+2]/[x_0+1]=1/[(x_0+1)^2](x-x_0)`
`<=>(-3x_0 -2)(x_0+1)=x-x_0`
`<=>-3x_0 ^2-3x_0 -2x_0 -2=x-x_0`
`<=>x=-3x_0 ^2-4x_0 -2`
`=>OA=|-3x_0 ^2-4x_0 -2|`
`@` Gọi `T T nn Oy=B`
`=>y-[3x_0 +2]/[x_0 +1]=1/[(x_0 +1)^2](-x_0)`
`<=>y=[(3x_0+2)(x_0+1)-x_0]/[(x_0+1)^2]`
`<=>y=[3x_0 ^2+4x_0 +2]/[(x_0 +1)^2]`
`=>OB=|[3x_0 ^2+4x_0 +2]/[(x_0 +1)^2]|`
Vì `\triangle OAB` vuông cân tại `O`
`=>OA=OB`
`<=>|-3x_0 ^2-4x_0 -2|=|[3x_0 ^2+4x_0 +2]/[(x_0 +1)^2]|`
`<=>(x_0+1)^2=1`
`<=>[(x_0=0),(x_0=-2):}`
`=>` PTTT: `[(y=x+2),(y=x+6):}`
\(\Leftrightarrow2sin^3x+1-sin^2x-1=0\)
\(\Leftrightarrow sin^2x\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=-cosx\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(x+\pi\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\pi+k2\pi\\2x+\dfrac{\pi}{6}=-x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{6}+k2\pi\\x=-\dfrac{7\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
c.
\(\Leftrightarrow2cos4x.sin3x=2sin4x.cos4x\)
\(\Leftrightarrow cos4x\left(sin4x-sin3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=0\\sin4x=sin3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k\pi\\4x=3x+k2\pi\\4x=\pi-3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=k2\pi\\x=\dfrac{\pi}{7}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
2.
\(f\left(x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x-5\)
\(=-\dfrac{9}{2}-\left(\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x\right)\)
\(=-\dfrac{9}{2}-cos\left(2x-\dfrac{\pi}{3}\right)\)
Do \(-1\le-cos\left(2x-\dfrac{\pi}{3}\right)\le1\Rightarrow-\dfrac{11}{2}\le y\le-\dfrac{7}{2}\)
\(y_{min}=-\dfrac{11}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=1\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
\(y_{max}=-\dfrac{7}{2}\) khi \(cos\left(2x-\dfrac{\pi}{3}\right)=-1\Rightarrow x=\dfrac{2\pi}{3}+k\pi\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-5x-6}{1-4x^3+x^2}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^3\left(3-\dfrac{5}{x^2}-\dfrac{6}{x^3}\right)}{x^3\left(\dfrac{1}{x^3}-4+\dfrac{1}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{5}{x^2}-\dfrac{6}{x^3}}{\dfrac{1}{x^3}-4+\dfrac{1}{x}}=\dfrac{3-0-0}{0-4+0}=-\dfrac{3}{4}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{\left(3x^2+8\right)\left(2x+1\right)}{5-4x^3}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(3+\dfrac{8}{x}\right)x\left(2+\dfrac{1}{x}\right)}{x^3\left(\dfrac{5}{x^3}-4\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(3+\dfrac{8}{x}\right)\left(2+\dfrac{1}{x}\right)}{\dfrac{5}{x^3}-4}=\dfrac{\left(3+0\right)\left(2+0\right)}{0-4}=-\dfrac{6}{4}=-\dfrac{3}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{-5x+7}{3-2x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(-5+\dfrac{7}{x}\right)}{x\left(\dfrac{3}{x}-2\right)}=\lim\limits_{x\rightarrow+\infty}\dfrac{-5+\dfrac{7}{x}}{\dfrac{3}{x}-2}=\dfrac{-5+0}{0-2}=\dfrac{5}{2}\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{7}{2x-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{7}{x}}{2-\dfrac{1}{x}}=\dfrac{0}{2-0}=0\)
2.
\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)
\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)
\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)
\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)
\(\Leftrightarrow cosx\left(cosx+sinx\right)\left(cosx-sinx-1\right)=0\)
\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x=k2\pi\end{matrix}\right.\)
a.
\(SH\perp\left(ABC\right)\Rightarrow SH=d\left(S;\left(ABC\right)\right)\)
\(SH\perp\left(ABC\right)\Rightarrow SH\perp BC\Rightarrow\Delta SBH\) vuông tại H
\(BH=\dfrac{1}{2}BC=a\Rightarrow SH=\sqrt{SB^2-BH^2}=a\sqrt{3}\)
\(SH\perp\left(ABC\right)\Rightarrow HA\) là hình chiếu vuông góc của SA lên (ABC)
\(\Rightarrow\widehat{SAH}\) là góc giữa SA và (ABC)
\(AH=\dfrac{1}{2}BC=a\) (trung tuyến ứng với cạnh huyền)
\(\Rightarrow tan\widehat{SAH}=\dfrac{SH}{AH}=\sqrt{3}\Rightarrow\widehat{SAH}=60^0\)
b.
H là trung điểm BC, M là trung điểm AB \(\Rightarrow MH\) là đường trung bình tam giác ABC
\(\Rightarrow MH||AC\Rightarrow MH\perp AB\) (do \(AB\perp AC\))
Lại có \(SH\perp\left(ABC\right)\Rightarrow SH\perp AB\)
\(\Rightarrow AB\perp\left(SMH\right)\)
Mà \(AB=\left(SAB\right)\cap\left(ABC\right)\Rightarrow\widehat{SMH}\) là góc giữa (SAB) và (ABC)
\(AC=\sqrt{BC^2-AB^2}=a\sqrt{3}\) \(\Rightarrow MH=\dfrac{1}{2}AC=\dfrac{a\sqrt{3}}{2}\) (đường trung bình)
\(\Rightarrow tan\widehat{SMH}=\dfrac{SH}{MH}=2\Rightarrow\widehat{SMH}\approx63^023'\)
c.
Theo cmt: \(\left\{{}\begin{matrix}MH\perp SH\\MH\perp AB\end{matrix}\right.\) \(\Rightarrow MH\) là đường vuông góc chung của SH và AB
\(\Rightarrow d\left(SH;AB\right)=MH=\dfrac{a\sqrt{3}}{2}\)
Từ H kẻ HK vuông góc SM (K thuộc SM)
\(AB\perp\left(SMH\right)\Rightarrow AB\perp HK\)
\(\Rightarrow HK\perp\left(SAB\right)\Rightarrow HK=d\left(H;\left(SAB\right)\right)\)
Hệ thức lượng trong tam giác vuông SMH:
\(HK=\dfrac{SH.MH}{\sqrt{SH^2+MH^2}}=\dfrac{a\sqrt{15}}{5}\)