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1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
3.
\(SA\perp\left(ABC\right)\Rightarrow\widehat{SBA}\) là góc giữa SB và (ABC)
\(AB=\sqrt{AC^2+BC^2}=a\sqrt{3}\)
\(tan\widehat{SBA}=\frac{SA}{AB}=\frac{1}{\sqrt{3}}\Rightarrow\widehat{SBA}=30^0\)
4.
\(f'\left(x\right)=\frac{\left(x^2+3\right)'}{2\sqrt{x^2+3}}=\frac{x}{\sqrt{x^2+3}}\) \(\Rightarrow\left\{{}\begin{matrix}f\left(1\right)=2\\f'\left(1\right)=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow S=2+4.\frac{1}{2}=4\)
5.
Hàm \(y=\frac{3}{x^2+2}\) xác định và liên tục trên R
6.
\(\left\{{}\begin{matrix}k_1=f'\left(2\right)\\k_2=g'\left(2\right)\\k_3=\frac{f'\left(2\right).g\left(2\right)-g'\left(2\right).f\left(2\right)}{g^2\left(2\right)}\end{matrix}\right.\) \(\Rightarrow k_3=\frac{k_1.g\left(2\right)-k_2.f\left(2\right)}{g^2\left(2\right)}\Rightarrow\frac{1}{2}=\frac{g\left(2\right)-f\left(2\right)}{g^2\left(2\right)}\)
\(\Leftrightarrow g^2\left(2\right)=2g\left(2\right)-2f\left(2\right)\)
\(\Leftrightarrow1-2f\left(2\right)=\left[g\left(2\right)-1\right]^2\ge0\)
\(\Rightarrow2f\left(2\right)\le1\Rightarrow f\left(2\right)\le\frac{1}{2}\)
1.
\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\Rightarrow SA\perp BC\\BC\perp AB\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)
\(\Rightarrow d\left(C;\left(SAB\right)\right)=BC\)
\(BC=\sqrt{AC^2-AB^2}=a\)
2.
Qua S kẻ đường thẳng d song song AD
Kéo dài AM cắt d tại E \(\Rightarrow SADE\) là hình chữ nhật
\(\Rightarrow DE//SA\Rightarrow ED\perp\left(ABCD\right)\)
\(SBCE\) cũng là hcn \(\Rightarrow SB//CE\Rightarrow SB//\left(ACM\right)\Rightarrow d\left(SB;\left(ACM\right)\right)=d\left(B;\left(ACM\right)\right)\)
Gọi O là tâm đáy, BD cắt (ACM) tại O, mà \(BO=DO\)
\(\Rightarrow d\left(B;\left(ACM\right)\right)=d\left(D;\left(ACM\right)\right)\)
\(\left\{{}\begin{matrix}AC\perp BD\\AC\perp ED\end{matrix}\right.\) \(\Rightarrow AC\perp\left(BDE\right)\)
Từ D kẻ \(DH\perp OE\Rightarrow DH\perp\left(ACM\right)\Rightarrow DH=d\left(D;\left(ACM\right)\right)\)
\(BD=a\sqrt{2}\Rightarrow OD=\frac{1}{2}BD=\frac{a\sqrt{2}}{2}\) ; \(ED=SA=2a\)
\(\frac{1}{DH^2}=\frac{1}{DO^2}+\frac{1}{ED^2}=\frac{9}{4a^2}\Rightarrow DH=\frac{2a}{3}\)
a/ \(\frac{A^4_n}{A_{n+1}^3-C_n^{n-4}}=\frac{24}{23}\Rightarrow n=5\)
Khai triển \(\left(2-3x^2+x^3\right)^5\)
\(\left\{{}\begin{matrix}k_0+k_2+k_3=5\\2k_2+3k_3=9\end{matrix}\right.\) \(\Rightarrow\left(k_0;k_2;k_3\right)=\left(1;3;1\right);\left(2;0;3\right)\)
Hệ số của số hạng chứa \(x^9\):
\(\frac{5!}{1!.3!.1!}.2^1.\left(-3\right)^3+\frac{5!}{2!.3!}.2^2.\left(-3\right)^0=-1040\)
b/ SHTQ của khai triển: \(\left(1+2x\right)^n\) là: \(C_n^k2^kx^k\)
\(\Rightarrow\) Hệ số của \(x^3\) trong khai triển tổng quát là \(C_n^32^3\)
\(\Rightarrow\) Hệ số của \(x^3\) trong khai triển của \(f\left(x\right)\): \(2^3.\sum\limits^{22}_{n=3}C_n^3\)
Tính tổng \(C_3^3+C_4^3+C_5^3+...+C_{22}^3\)
\(=C_4^4+C_4^3+C_5^3+...+C_{22}^3\)
\(=C_5^4+C_5^3+...+C_{22}^3\)
\(=C_6^4+C_6^3+...+C_{22}^3=...=C_{23}^4\)
Vậy \(2^3\sum\limits^{22}_{n=3}C_n^3=2^3.C_{23}^4\)
Lời giải khác:
Theo BĐT AM-GM:
\(\text{VT}=\sum \frac{\sqrt{2(b^2+c^2)-a^2}}{a}\geq \sum \frac{\sqrt{(b+c)^2-a^2}}{a}=\sum \frac{\sqrt{a+b+c}.\sqrt{b+c-a}}{a}\)
\(=\sum \frac{\sqrt{a+b+c}.(b+c-a)}{\sqrt{a^2(b+c-a)}}\)
Theo BĐT AM-GM:
$a^2(b+c-a)\leq \left(\frac{a+b+c}{3}\right)^3$
\(\Rightarrow \text{VT}\geq 3\sqrt{3}\sum \frac{\sqrt{a+b+c}(b+c-a)}{\sqrt{(a+b+c)^3}}=3\sqrt{3}.\sum \frac{b+c-a}{a+b+c}=3\sqrt{3}\)
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Chuẩn hóa \(a+b+c=3\)
Do a;b;c là độ dài 3 cạnh của 1 tam giác nên ta cũng suy ra \(0< a;b;c< \frac{3}{2}\)
Đặt vế trái là P, ta có:
\(P=\sum\frac{\sqrt{2\left(b^2+c^2\right)-a^2}}{a}\ge\sum\frac{\sqrt{\left(b+c\right)^2-a^2}}{a}=\sum\frac{\sqrt{\left(a+b+c\right)\left(b+c-a\right)}}{a}=\sqrt{3}\left(\frac{\sqrt{3-2a}}{a}+\frac{\sqrt{3-2b}}{b}+\frac{\sqrt{3-2c}}{c}\right)\)
Ta có đánh giá: \(\frac{\sqrt{3-2a}}{a}\ge3-2a\) với mọi \(a\in\left(0;\frac{3}{2}\right)\)
Thật vậy, BĐT \(\Leftrightarrow a\sqrt{3-2a}\le1\)
\(\Leftrightarrow1-a^2\left(3-2a\right)\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(2a+1\right)\ge0\) (luôn đúng)
Tương tự \(\frac{\sqrt{3-2b}}{b}\ge3-2b\) ; \(\frac{\sqrt{3-2c}}{c}\ge3-2c\)
\(\Rightarrow P\ge\sqrt{3}\left[9-2\left(a+b+c\right)\right]=3\sqrt{3}\) (đpcm)
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Câu hỏi của Thảo Nguyễn Phương - Toán lớp 11 | Học trực tuyến