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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: \(S_{ABCD}=AB^2=a^2\)
\(V_{S.ABCD}=\dfrac{1}{3}\cdot SA\cdot S_{ABCD}=\dfrac{1}{3}\cdot3a\cdot a^2=a^3\)
Câu 1: \(a\cdot\sqrt[3]{a}=a\cdot a^{\dfrac{1}{3}}=a^{\dfrac{4}{3}}\)
=>Chọn C
Câu 2:
ĐKXĐ: x+3>0
=>x>-3
=>Chọn C
Câu 3:
\(3^{x+2}=27\)
=>\(3^{x+2}=3^3\)
=>x+2=3
=>x=1
Câu 4:
ĐKXĐ: x>0
\(log_2^2x-5\cdot log_2x-6< =0\)
=>\(\left(log_2x-6\right)\left(log_2x+1\right)< =0\)
=>\(log_2x-6< =0\)
=>\(log_2x< =6\)
=>x<=64
=>0<x<=64
=>Chọn B
Câu 9:
\(P\left(AB\right)=0,7\cdot0,2=0,14\)
=>Chọn A
Câu 9:
\(P\left(\overline{A}\right)=1-0,4=0,6\)
\(P\left(\overline{A}B\right)=0,6\cdot0,5=0,3\)
=>Chọn B
Câu 10:
A: "Tổng số chấm trên hai con xúc sắc là 5"
=>A={(1;4);(2;3);(3;2);(4;1)}
B: "Tích số chấm trên hai con xúc sắc là 6"
=>B={(1;6);(6;1);(2;3);(3;2)}
=>\(A\cap B=\left\{\left(2;3\right);\left(3;2\right)\right\}\)
=>Chọn D
Câu 11:
A: "Tổng số chấm trên hai con xúc sắc là 7"
=>A={(1;6);(2;5);(5;2);(6;1);(3;4);(4;3)}
B: "Tích số chấm trên hai con xúc sắc là 10"
=>B={(2;5);(5;2)}
=>\(A\cap B=\left\{\left(2;5\right);\left(5;2\right)\right\}\)
=>Chọn A
Câu 11:
\(f\left(x\right)=2x+cosx\)
=>\(f'\left(x\right)=2-sinx\)
\(-1< =-sinx< =1\)
=>\(-1+2< =f\left(x\right)< =1+2\)
=>1<=f(x)<=3
=>Chọn B
Câu 12:
\(y=x^3-3x^2+2\)
=>\(y'=3x^2-3\cdot2x=3x^2-6x\)
\(y'\left(-1\right)=3\cdot\left(-1\right)^2-6\cdot\left(-1\right)=3+6=9\)
\(y\left(-1\right)=\left(-1\right)^3-3\cdot\left(-1\right)^2+2=-1+2-3=-4+2=-2\)
Phương trình tiếp tuyến tại x=-1 là:
y-y(-1)=y'(-1)(x+1)
=>y-(-2)=9(x+1)
=>y+2=9x+9
=>y=9x+7
=>Chọn B