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1a.
\(y'=3x^2.f'\left(x^3\right)-2x.g'\left(x^2\right)\)
b.
\(y'=\dfrac{3f^2\left(x\right).f'\left(x\right)+3g^2\left(x\right).g'\left(x\right)}{2\sqrt{f^3\left(x\right)+g^3\left(x\right)}}\)
2.
\(f'\left(x\right)=\left(m-1\right)x^3+\left(m-2\right)x^2-2mx+3=0\)
Để ý rằng tổng hệ số của vế trái bằng 1 nên pt luôn có nghiệm \(x=1\), sử dụng lược đồ Hooc-ne ta phân tích được:
\(\Leftrightarrow\left(x-1\right)\left[\left(m-1\right)x^2+\left(2m-3\right)x-3\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(m-1\right)x^2+\left(2m-3\right)x-3=0\left(1\right)\end{matrix}\right.\)
Xét (1), với \(m=1\Rightarrow x=-3\)
- Với \(m\ne1\Rightarrow\Delta=\left(2m-3\right)^2+12\left(m-1\right)=4m^2-3\)
Nếu \(\left|m\right|< \dfrac{\sqrt{3}}{2}\Rightarrow\) (1) vô nghiệm \(\Rightarrow f'\left(x\right)=0\) có đúng 1 nghiệm
Nếu \(\left|m\right|>\dfrac{\sqrt{3}}{2}\Rightarrow\left(1\right)\) có 2 nghiệm \(\Rightarrow f'\left(x\right)=0\) có 3 nghiệm
1/ \(y'=\dfrac{\left(\sqrt{x+1}\right)'x-x'\sqrt{x+1}}{x^2}=\dfrac{\dfrac{x}{2\sqrt{x+1}}-\sqrt{x+1}}{x^2}=\dfrac{-x-2}{2x^2\sqrt{x+1}}\)
2/ \(y'=\dfrac{1-x^2-\left(1-x^2\right)'x}{\left(1-x^2\right)^2}=\dfrac{1+x^2}{\left(1-x^2\right)^2}\)
3/ \(y'=\dfrac{-\left(x-\sqrt{x+1}\right)'}{\left(x-\sqrt{x+1}\right)^2}=\dfrac{-1+\dfrac{1}{2\sqrt{x+1}}}{\left(x-\sqrt{x+1}\right)^2}\)
4/ \(y'=f'\left(x\right)=2x-\dfrac{2x}{x^4}=2x-\dfrac{2}{x^3}\)
\(y'=0\Leftrightarrow\dfrac{2x^4-2}{x^3}=0\Leftrightarrow x=\pm1\)
5/ \(y'=\dfrac{\dfrac{1}{2\sqrt{1+x}}}{2\sqrt{1+\sqrt{1+x}}}\Rightarrow f\left(x\right).f'\left(x\right)=\sqrt{1+\sqrt{1+x}}.\dfrac{1}{4\sqrt{1+x}.\sqrt{1+\sqrt{1+x}}}=\dfrac{1}{4\sqrt{1+x}}=\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow2\sqrt{1+x}=\sqrt{2}\Leftrightarrow1+x=\dfrac{1}{2}\Leftrightarrow x=-\dfrac{1}{2}\)
Hãy nhớ câu tính đạo hàm này, bởi nó liên quan đến nguyên hàm sau này sẽ học
1. Áp dụng quy tắc L'Hopital
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1}{f\left(0\right)-f\left(x\right)}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2\sqrt{x+1}}}{-f'\left(0\right)}=-\dfrac{1}{6}\)
2.
\(g'\left(x\right)=2x.f'\left(\sqrt{x^2+4}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\f'\left(\sqrt{x^2+4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x^2+4}=1\\\sqrt{x^2+4}=-2\end{matrix}\right.\)
2 pt cuối đều vô nghiệm nên \(g'\left(x\right)=0\) có đúng 1 nghiệm
1/ L'Hospital:
\(=\lim\limits_{x\rightarrow6}f'\left(x\right)=f'\left(6\right)=2\)
3/ \(=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{3}{2\sqrt{3x+3}}}{1}=\dfrac{1}{2}\Rightarrow2a-b=0\)
4/ \(=\lim\limits_{x\rightarrow1}\dfrac{2f\left(x\right).f'\left(x\right)-f'\left(x\right)}{\dfrac{1}{2\sqrt{x}}}=\dfrac{2.6.5-5}{\dfrac{1}{2}}=110\)
2/ \(x_0=-3\Rightarrow y_0=\dfrac{-3-1}{-3+2}=\dfrac{-4}{-1}=4\)
\(y'=\dfrac{\left(x-1\right)'\left(x+2\right)-\left(x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}=\dfrac{x+2-x+1}{\left(x+2\right)^2}=\dfrac{3}{\left(x+2\right)^2}\)
\(\Rightarrow y'\left(-3\right)=3\)
\(\Rightarrow pttt:y=3\left(x+3\right)+4=3x+13\)
\(x=0\Rightarrow y=13;y=0\Rightarrow x=-\dfrac{13}{3}\)
\(\Rightarrow S=\dfrac{1}{2}.\left|x\right|\left|y\right|=\dfrac{1}{2}.\dfrac{13}{3}.13=\dfrac{169}{6}\left(dvdt\right)\)
P/s: Câu 5,6 bỏ qua nhé, toi ngu hình học :b
`f'(x) = x^2 - 4x+m`
`f'(x) >=0 <=>x^2-4x+m>=0`
`<=> \Delta' >=0`
`<=> 2^2-1.m>=0`
`<=> m<=4`
Vậy....
\(f'=6x^8-6x^5+6x+6=6\left(x^8-x^5+x+1\right)\)
\(\left[{}\begin{matrix}\left|x\right|\le1\Rightarrow\left|x^5-x\right|\le\left|x\right|\le1\Rightarrow1-x^5-x\ge0\\\left|x\right|\ge1\Rightarrow\left|x^5\right|\le x^8\Rightarrow\left\{{}\begin{matrix}x^8-x^5>0\\x^2-x>0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow f'\left(x\right)>0\forall x\)
\(f'\left(x\right)=\dfrac{\left(x^2\right)'\left(x+1\right)-x^2\left(x+1\right)'}{\left(x+1\right)^2}\)
\(=\dfrac{2x\left(x+1\right)-x^2}{\left(x+1\right)^2}=\dfrac{x^2+2x}{x^2+2x+1}\)
\(f'\left(x\right)=\dfrac{\left(2x-3\right)\left(x-1\right)-x^2+3x-7}{\left(x-1\right)^2}=\dfrac{x^2-2x-4}{\left(x-1\right)^2}\)
\(f'\left(x\right)=0\Leftrightarrow x^2-2x-4=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow x_1^2+x_2^2=12\)
Hoặc bạn dùng Vi-ét cũng được, tùy
\(f'\left(x\right)=\dfrac{\left(x^2\right)'\cdot\left(x+1\right)-x^2\cdot\left(x+1\right)'}{\left(x+1\right)^2}\)
\(=\dfrac{2x\left(x+1\right)-x^2}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
\(y'=\dfrac{x'\left(x+1\right)-x\left(x+1\right)'}{\left(x+1\right)^2}=\dfrac{x+1-x}{\left(x+1\right)^2}=\dfrac{1}{\left(x+1\right)^2}\)
\(y'\left(0\right)=\dfrac{1}{\left(0+1\right)^2}=1\)