Giúp e làm chi tiết câu 21, 22 đi ạ
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\(\lim\limits_{x\rightarrow+\infty}\left(ax-\sqrt{x^2+bx+2}\right)=\lim\limits_{x\rightarrow+\infty}x\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)\)
Nếu \(a\ne1\Rightarrow\lim\limits_{x\rightarrow+\infty}\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)=a-1\ne0\)
\(\Rightarrow\lim\limits_{x\rightarrow+\infty}x\left(a-\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}\right)=\infty\) ko thỏa mãn giả thiết \(=4\) (hữu hạn)
\(\Rightarrow a=1\)
\(\lim\limits_{x\rightarrow+\infty}\left(x-\sqrt{x^2+bx+2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{-bx-2}{x+\sqrt{x^2+bx+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-b-\dfrac{2}{x}}{1+\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}}=-\dfrac{b}{2}\)
\(\Rightarrow-\dfrac{b}{2}=4\Rightarrow b=-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
21:
\(y'=\dfrac{\left(x^2-3x+5\right)'\left(x+2\right)-\left(x+2\right)'\left(x^2-3x+5\right)}{\left(x+2\right)^2}\)
\(=\dfrac{\left(2x-3\right)\left(x+2\right)-\left(x^2-3x+5\right)}{\left(x+2\right)^2}\)
\(=\dfrac{2x^2+4x-3x-6-x^2+3x-5}{\left(x+2\right)^2}=\dfrac{x^2+4x-11}{\left(x+2\right)^2}\)
17:
Khi x<>0 thì \(\lim\limits_{x\rightarrow0}f\left(x\right)=\lim\limits_{x\rightarrow0}\dfrac{1+4x-1}{\sqrt{1+4x}+1}\cdot\dfrac{1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{4}{\sqrt{1+4x}+1}=\dfrac{4}{1+1}=\dfrac{4}{2}=2\)
=>Chọn B
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\(f\left(x\right)=x^5+x^3\Rightarrow f'\left(x\right)=5x^4+3x^2\)
\(f'\left(2\right)=5.2^4+3.2^2=92\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5.
\(AA'\perp\left(A'B'C'D'\right)\) theo t/c lập phương
\(\Rightarrow AA'\perp B'C'\Rightarrow\) góc giữa 2 đường thẳng bằng 90 độ
6.
\(y'=\left(x.cosx\right)'=x'.cosx+\left(cosx\right)'.x=cosx-x.sinx\)
7.
\(y'=-3x^2-5\)
\(y''=-6x\)
8.
\(\lim\limits_{x\rightarrow+\infty}\left(x^3+3x-2\right)=\lim\limits_{x\rightarrow+\infty}x^3\left(1+\dfrac{3}{x}-\dfrac{2}{x^3}\right)=+\infty.1=+\infty\)
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Câu 5:
\(\Leftrightarrow-x^2+7x-9+2x-9=0\)
\(\Leftrightarrow x^2-9x+18=0\)
=>x=3
=>Chọn A
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo tính chất hình lập phương, ta có:
\(C'D'\perp\left(BB'C'C\right)\Rightarrow C'D'\perp BC'\)
\(\Rightarrow\widehat{\left(C'D';BC'\right)}=90^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y'=\left(3x^2\right)'-\left(4x\right)'+9'\)
\(y'=6x-4\Rightarrow y'\left(1\right)=6.1-4=2\)
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\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{mx^2-\left(m+3\right)x+3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(mx-3\right)}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(mx-3\right)=m-3\)
\(f\left(1\right)=m^2-15\)
Hàm liên tục tại \(x=1\) khi:
\(m-3=m^2-15\Rightarrow m^2-m-12=0\Rightarrow\left[{}\begin{matrix}m=4\\m=-3\end{matrix}\right.\)
\(4^2+\left(-3\right)^2=25\)
21.
Giới hạn đã cho hữu hạn khi và chỉ khi \(a=1\)
Khi đó:
\(\lim\limits_{x\rightarrow+\infty}\left(x-\sqrt{x^2+bx+2}\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2-\left(x^2+bx+2\right)}{x+\sqrt{x^2+bx+2}}=\lim\limits_{x\rightarrow+\infty}\dfrac{-bx-2}{x+\sqrt{x^2+bx+2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-b-\dfrac{2}{x}}{1+\sqrt{1+\dfrac{b}{x}+\dfrac{2}{x^2}}}=\dfrac{-b}{2}\)
\(\Rightarrow-\dfrac{b}{2}=4\Rightarrow b=-8\)
\(\Rightarrow a+b=1-8=-7\)
22.
B sai, do các cạnh bên của chóp đều tạo với đáy các góc bằng nhau