1: \(y=2x+cosx\)

=>\(y'=2-sinx\)

=>\(y''=2'-\left(sinx\right)'=-cosx\)

2: \(y=sin^3x\)

=>\(y'=3\cdot sin^2x\cdot\left(sinx\right)'=3\cdot sin^2x\cdot cosx\)

=>\(y''=3\cdot\left(sin^2x\cdot cosx\right)'\)

=>\(y''=3\left[\left(sin^2x\right)'\cdot cosx+\left(sin^2x\right)\cdot\left(cosx\right)'\right]\)

=>\(y''=3\left[2\cdot sinx\cdot\left(sinx\right)'\cdot cosx+sin^2x\cdot\left(-sinx\right)\right]\)

=>\(y''=3\left[2\cdot sinx\cdot cosx\cdot sinx-sin^3x\right]\)

=>\(y''=6\cdot sin^2x\cdot cosx-3\cdot sin^3x\)

3: \(y=2\cdot sin2x-cos\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y'=2\cdot\left(2x\right)'\cdot\left(cos2x\right)-\left(-1\right)\cdot\left(x+\dfrac{\Omega}{3}\right)'\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y'=4\cdot cos2x+sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y''=4\cdot\left(-1\right)\cdot\left(2x\right)'\cdot sin2x+\left(x+\dfrac{\Omega}{3}\right)'\cdot cos\left(x+\dfrac{\Omega}{3}\right)\)

=>\(y''=-8\cdot sin2x+cos\left(x+\dfrac{\Omega}{3}\right)\)

4: \(y=\sqrt{x^2+1}\)

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

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

=>\(y''=\dfrac{\sqrt{x^2+1}-x\cdot\dfrac{x}{\sqrt{x^2+1}}}{x^2+1}\)

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

5: \(y=x\cdot cosx\)

=>\(y'=x'\cdot cosx+x\cdot\left(cosx\right)'=cosx-sinx\cdot x\)

=>\(y''=\left(cosx\right)'-\left(sinx\cdot x\right)'\)

=>\(y''=-sinx-\left[\left(sinx\right)'\cdot x+sinx\cdot x'\right]\)

=>\(y''=-sinx-cosx\cdot x-sinx\)

=>\(y''=-2\cdot sinx-cosx\cdot x\)

6: \(y=\dfrac{x+2}{x-3}\)

=>\(y'=\dfrac{\left(x+2\right)'\left(x-3\right)-\left(x+2\right)\left(x-3\right)'}{\left(x-3\right)^2}\)

=>\(y''=\dfrac{x-3-x-2}{\left(x-3\right)^2}=\dfrac{-5}{\left(x-3\right)^2}\)

=>\(y''=\dfrac{\left(-5\right)'\cdot\left(x-3\right)^2-\left(-5\right)\cdot\left[\left(x-3\right)^2\right]'}{\left(x-3\right)^4}\)

=>\(y''=\dfrac{5\cdot\left(x^2-6x+9\right)'}{\left(x-3\right)^4}\)

=>\(y''=\dfrac{5\left(2x-6\right)}{\left(x-3\right)^4}=\dfrac{10}{\left(x-3\right)^3}\)

\(u_{n+1}=u_n+\dfrac{1}{n\left(n+1\right)}\Rightarrow u_{n+1}=u_n+\dfrac{1}{n}-\dfrac{1}{n+1}\)

\(\Rightarrow u_{n+1}+\dfrac{1}{n+1}=u_n+\dfrac{1}{n}\)

Đặt \(u_n+\dfrac{1}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1+\dfrac{1}{1}=2\\v_{n+1}=v_n\end{matrix}\right.\)

Từ \(v_{n+1}=v_n\Rightarrow v_{n+1}=v_n=v_{n-1}=...=v_1=2\)

\(\Rightarrow v_n=2\Rightarrow u_n+\dfrac{1}{n}=2\)

\(\Rightarrow u_n=2-\dfrac{1}{n}=\dfrac{2n-1}{n}\)

\(\Rightarrow u_{2024}=\dfrac{2.2024-1}{2024}=\dfrac{4047}{2024}\)

a.

\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\AD\perp CD\left(gt\right)\end{matrix}\right.\)

\(\Rightarrow CD\perp\left(SAD\right)\)

b.

\(V=\dfrac{1}{3}SA.AB.AD=2a^3\)

Câu 3:

\(P\left(AB\right)=P\left(A\right)\cdot P\left(B\right)=0,9\cdot0,7=0,63\)

\(P\left(\overline{A}\right)=1-0,9=0,1;P\left(\overline{B}\right)=1-0,7=0,3\)

\(P\left(\overline{A}B\right)=0,1\cdot0,7=0,07\)

\(P\left(\overline{A}\overline{B}\right)=0,1\cdot0,3=0,03\)

Câu 1:

\(y=x^4-x+1\)

=>\(y'=4x^3-1\)

\(y'\left(2\right)=4\cdot2^3-1=4\cdot8-1=31\)

Phương trình tiếp tuyến tại M là:

y-y(2)=y'(2)(x-2)

=>y-15=31(x-2)

=>y-15=31x-62

=>y=31x-62+15=31x-47

\(P\left(AB\right)=P\left(A\right).P\left(B\right)=0,9.0,7=0,63\)

\(P\left(\overline{A}B\right)=P\left(B\right)-P\left(AB\right)=0,7-0,63=0,07\)

\(P\left(\overline{AB}\right)=1-P\left(AB\right)=0,37\)

Gọi M là trung điểm EG \(\Rightarrow AM\perp EG\) (tam giác cân)

\(\Rightarrow AM\perp\left(EFGH\right)\Rightarrow AM=d\left(A;\left(EFGH\right)\right)\)

\(EG=30-2x\Rightarrow EM=\dfrac{1}{2}EG=15-x\)

\(\Rightarrow AM=\sqrt{AE^2-EM^2}=\sqrt{x^2-\left(15-x\right)^2}=\sqrt{30x-225}\)

Do AEG là tam giác, theo BĐT tam giác: \(\left\{{}\begin{matrix}AE+AG>EG\\\left|AG-AE\right|< EG\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+x>30-2x\\0< 30-2x\end{matrix}\right.\) \(\Rightarrow\dfrac{15}{2}< x< 15\)

\(V=AD.S_{\Delta AEG}=30.\dfrac{1}{2}AM.EG=15.\left(30-2x\right)\sqrt{30x-225}\)

\(V^2=15^3.4\left(15-x\right)^2\left(2x-15\right)=15^3.4.\left(15-x\right)\left(15-x\right)\left(2x-15\right)\)

\(\le15^3.4.\left(\dfrac{15-x+15-x+2x-15}{3}\right)^3=...\)

Dấu "=" xảy ra khi \(15-x=2x-15\Rightarrow x=10\)

\(\Rightarrow d\left(A;\left(EFGH\right)\right)=AM=\sqrt{30.10-225}=5\sqrt{3}\)

a/

Ta có

ABCD là HCN (gt) \(\Rightarrow BC\perp AB\)

\(SA\perp\left(ABCD\right);BC\in\left(ABCD\right)\Rightarrow SA\perp BC\)

\(\Rightarrow BC\perp AB;BC\perp SA\Rightarrow BC\perp\left(SAB\right)\) mà \(SB\in\left(SBC\right)\)

\(\Rightarrow BC\perp SB\) => tg SBC vuông tại B

b/ Chứng minh tương tự cũng có

\(CD\perp\left(SAD\right)\) mà \(SD\in\left(SCD\right)\)

\(\Rightarrow CD\perp SD\) => tg SCD vuông tại D

c/

Ta có

\(BC\perp\left(SAB\right)\left(cmt\right);AH\in\left(SAB\right)\Rightarrow AH\perp BC\) 

mà \(AH\perp SB\left(gt\right)\)

\(\Rightarrow AH\perp\left(SBC\right)\) mà \(SC\in\left(SBC\right)\)

\(\Rightarrow SC\perp AH\)

C/m tương tự ta cũng có \(SC\perp AK\)

\(\Rightarrow SC\perp\left(AHK\right)\)

d/

Ta có

\(SC\perp\left(AHK\right)\left(cmt\right);HK\in\left(AHK\right)\Rightarrow HK\perp SC\)

1.

\(y'=\dfrac{2x+4}{2\sqrt{x^2+4x+3}}=\dfrac{x+2}{\sqrt{x^2+4x+3}}\)

2.

\(f'\left(x\right)=\dfrac{1}{2\sqrt{x+1}}+\dfrac{1}{2\sqrt{x-1}}\)

3.

\(y'=\dfrac{\left(2x+2\right)\left(x-2\right)-\left(x^2+2x+1\right)}{\left(x-2\right)^2}=\dfrac{x^2-4x-5}{\left(x-2\right)^2}\)

4.

\(y'=\dfrac{-6}{\left(6x-5\right)^2}\)

5.

\(y'=\dfrac{2.\left(-1\right)-1.1}{\left(x-1\right)^2}=\dfrac{-3}{\left(x-1\right)^2}\)

6.

\(y'=4x^3+4x\)

Câu 1:

\(y=x^3+2x^2+1\)

=>\(y'=3x^2+2\cdot2x=3x^2+4x\)

\(y'\left(1\right)=3\cdot1^2+4\cdot1=3+4=7\)

Phương trình tiếp tuyến tại x=1 là:

y-f(1)=f'(1)(x-1)

=>y-4=7(x-1)

=>y=7x-7+4=7x-3

Câu 2:

a: BD\(\perp\)AC(ABCD là hình vuông)

BD\(\perp\)SA(SA\(\perp\)(ABCD))

AC,SA cùng thuộc mp(SAC)

Do đó: BD\(\perp\)(SAC)

b: \(S_{ABCD}=AB^2=a^2\)

\(V_{S.ABCD}=\dfrac{1}{3}\cdot SA\cdot S_{ABCD}=\dfrac{1}{3}\cdot3a\cdot a^2=a^3\)

a.

\(S_{\Delta ABC}=\dfrac{1}{2}AB.AC.sinA=\dfrac{1}{2}.a.a.sin120^0=\dfrac{a^2\sqrt{3}}{4}\)

\(\Rightarrow V=\dfrac{1}{3}SA.S_{\Delta ABC}=\dfrac{a^3}{8}\)

b.

Gọi N là trung điểm AB \(\Rightarrow MN\) là đường trung bình tam giác ABC

\(\Rightarrow MN||AC\Rightarrow AC||\left(SMN\right)\)

\(\Rightarrow d\left(SM;AC\right)=d\left(AC;\left(SMN\right)\right)=d\left(A;\left(SMN\right)\right)\)

Từ A kẻ AH vuông góc MN (H thuộc đường thẳng MN)

Từ A kẻ \(AK\perp SH\)  (K thuộc SH) (1)

\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp MN\\AH\perp MN\left(gt\right)\end{matrix}\right.\) \(\Rightarrow MN\perp\left(SAH\right)\)

\(\Rightarrow MN\perp AK\) (2)

(1);(2)\(\Rightarrow AK\perp\left(SMN\right)\Rightarrow AK=d\left(A;\left(SMN\right)\right)\)

AH vuông góc MN, mà AC song song MN \(\Rightarrow AH\perp AC\Rightarrow\widehat{CAH}=90^0\)

\(\Rightarrow\widehat{HAN}=\widehat{BAC}-\widehat{CAH}=120^0-90^0=30^0\)

\(\Rightarrow AH=AN.cos\widehat{HAN}=\dfrac{AB}{2}.cos30^0=\dfrac{a\sqrt{3}}{4}\)

Hệ thức lượng:

\(AK=\dfrac{AH.SA}{\sqrt{AH^2+SA^2}}=\dfrac{a\sqrt{39}}{26}\)