no khó lắm tui lớp 5 thôi chịu

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1.a

\(\lim\limits_{x\rightarrow2}\dfrac{x^3+3x^2-9x-2}{x^3-x-6}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x^2+5x+1\right)}{\left(x-2\right)\left(x^2+2x+3\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{x^2+5x+1}{x^2+2x+3}=\dfrac{15}{11}\)

b.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2-x+3}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-x+3}{\sqrt{x^2-x+3}-x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{-1+\dfrac{3}{x}}{-\sqrt{1-\dfrac{1}{x}+\dfrac{3}{x^2}}-1}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

2.

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{2x-\sqrt{x^2+3}}{x-1}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{4x^2-\left(x^2+3\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x^2-1\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}\)

\(=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(2x+\sqrt{x^2+3}\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{3\left(x+1\right)}{2x+\sqrt{x^2+3}}\)

\(=\dfrac{3.2}{2+\sqrt{4}}=\dfrac{3}{2}\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\dfrac{2x+7}{6}=\dfrac{2.1+7}{6}=\dfrac{3}{2}\)

\(f\left(1\right)=\dfrac{2.1+7}{6}=\dfrac{3}{2}\)

\(\Rightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)

\(\Rightarrow\) Hàm liên tục tại \(x=1\)

3.

a.

\(y'=\dfrac{\left(2x^2-1\right)'\left(x-2\right)-\left(x-2\right)'\left(2x^2-1\right)}{\left(x-2\right)^2}\)

\(=\dfrac{4x\left(x-2\right)-2x^2+1}{\left(x-2\right)^2}=\dfrac{2x^2-8x+1}{\left(x-2\right)^2}\)

b.

\(y'=\left(\sqrt{1-2x^2}\right)'.\left(-sin\sqrt{1-2x^2}\right)\)

\(=-\dfrac{\left(1-2x^2\right)'}{2\sqrt{1-2x^2}}.sin\sqrt{1-2x^2}\)

\(=\dfrac{2x.sin\sqrt{1-2x^2}}{\sqrt{1-2x^2}}\)

c.

\(y'=\sqrt{3}+\sqrt{x}+x.\dfrac{1}{2\sqrt{x}}-\dfrac{2}{x^2}-5sinx\)

\(=\sqrt{3}+\dfrac{3\sqrt{x}}{2}-\dfrac{2}{x^2}-5sinx\)

d.

\(y'=\dfrac{4}{cos^24x}+sinx\)

4.

\(y'=\dfrac{\left(2-x+x^2\right)'\left(x-1\right)-\left(x-1\right)'\left(2-x+x^2\right)}{\left(x-1\right)^2}\)

\(=\dfrac{\left(-1+2x\right)\left(x-1\right)-\left(2-x+x^2\right)}{\left(x-1\right)^2}=\dfrac{x^2-2x-1}{\left(x-1\right)^2}\)

\(y'\left(2\right)=\dfrac{4-4-1}{\left(2-1\right)^2}=-1\)

Phương trình tiếp tuyến tại M có dạng:

\(y=-1\left(x-2\right)+4\Leftrightarrow y=-x+6\)

5.

\(f'\left(x\right)=\left(m-1\right)x^2-2x+1\)

Để \(f'\left(x\right)\ge0;\forall x\in R\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-1>0\\\Delta'=1-\left(m-1\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>1\\m\ge2\end{matrix}\right.\)

\(\Rightarrow m\ge2\)

6.

Ta có: \(\left\{{}\begin{matrix}SO\perp\left(ABCD\right)\Rightarrow SO\perp BD\\AC\perp BD\left(\text{hai đường chéo hv}\right)\end{matrix}\right.\)  \(\Rightarrow BD\perp\left(SAC\right)\)

Mà \(BD\in\left(SBD\right)\Rightarrow\left(SBD\right)\perp\left(SAC\right)\)

b.

\(SO\perp\left(ABCD\right)\Rightarrow SO\perp BC\) (1)

Lại có: O là trung điểm BD, M là trung điểm BC

\(\Rightarrow OM\) là đường trung bình tam giác BCD

\(\Rightarrow OM||CD\Rightarrow OM\perp BC\) (do \(BC\perp CD\)) (2)

(1);(2) \(\Rightarrow BC\perp\left(SOM\right)\)

c.

Từ O kẻ \(OE\perp SM\) (E thuộc SM)

\(\Rightarrow OE\perp\left(SBC\right)\)

\(\Rightarrow SE\) là hình chiếu vuông góc của SO lên (SBC)

\(\Rightarrow\widehat{OSE}\) là góc giữa SO và (SBC)

\(OM=\dfrac{1}{2}CD=\dfrac{a}{2}\)

\(\Rightarrow tan\widehat{OSE}=\dfrac{OM}{SO}=\dfrac{1}{2\sqrt{3}}\Rightarrow\widehat{OSE}\approx16^06'\)

Hình vẽ bài 6: