Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(y'=\dfrac{\left(-2x+2\right)\left(x-3\right)-\left(-x^2+2x+c\right)}{\left(x-3\right)^2}=\dfrac{-x^2+6x-6-c}{\left(x-3\right)^2}\)
\(\Rightarrow\) Cực đại và cực tiểu của hàm là nghiệm của: \(-x^2+6x-6-c=0\) (1)
\(\Delta'=9-\left(6+c\right)>0\Rightarrow c< 3\)
Gọi \(x_1;x_2\) là 2 nghiệm của (1) \(\Rightarrow\left\{{}\begin{matrix}-x_1^2+6x_1-6=c\\-x_2^2+6x_2-6=c\end{matrix}\right.\)
\(\Rightarrow m-M=\dfrac{-x_1^2+2x_1+c}{x_1-3}-\dfrac{-x_2^2+2x_2+c}{x_2-3}=4\)
\(\Leftrightarrow\dfrac{-2x_1^2+8x_1-6}{x_1-3}-\dfrac{-2x_2^2+8x_2-6}{x_2-3}=4\)
\(\Leftrightarrow2\left(1-x_1\right)-2\left(1-x_2\right)=4\)
\(\Leftrightarrow x_2-x_1=2\)
Kết hợp với Viet: \(\left\{{}\begin{matrix}x_2-x_1=2\\x_1+x_2=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2\\x_2=4\end{matrix}\right.\)
\(\Rightarrow c=2\)
Có 1 giá trị nguyên
bạn chỉ cần tách x4-1 thành (x2-1)(x2+1),rồi đặt x2=t là ok
Ta có: (u.v)' = u'.v + u.v'
\(Q=80K^{\dfrac{1}{3}}\left(100-K\right)^{\dfrac{1}{2}}\)
\(Q'=80.\left(K^{\dfrac{1}{3}}\right)'.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\left(\left(100-K\right)^{\dfrac{1}{2}}\right)'\)= \(80.\dfrac{1}{3}.K^{-\dfrac{2}{3}}.\left(100-K\right)^{\dfrac{1}{2}}+80.K^{\dfrac{1}{3}}.\dfrac{1}{2}.\left(100-K\right)^{-\dfrac{1}{2}}.\left(-1\right)\) = \(80.\left(\dfrac{\left(100-K\right)^{\dfrac{1}{2}}}{3K^{\dfrac{2}{3}}}-\dfrac{K^{\dfrac{1}{3}}}{2\left(100-K\right)^{\dfrac{1}{2}}}\right)\)= \(80.\left(\dfrac{2\left(100-K\right)^{\dfrac{1}{2}}\left(100-K\right)^{\dfrac{1}{2}}-3K^{\dfrac{2}{3}}K^{\dfrac{1}{3}}}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{2\left(100-K\right)-3K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(80.\left(\dfrac{200-5K}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\right)\) = \(\dfrac{400\left(40-K\right)}{6K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\) = \(\dfrac{200\left(40-K\right)}{3K^{\dfrac{2}{3}}\left(100-K\right)^{\dfrac{1}{2}}}\).
bạn tải về rồi zoom lên ý, vì đây là tớ chụp ảnh nên ảnh nhỏ
mong bạn tải về zoom lên hướng dẫn tớ với
\(I=\int\dfrac{2}{2+5sinxcosx}dx=\int\dfrac{2sec^2x}{2sec^2x+5tanx}dx\\ =\int\dfrac{2sec^2x}{2tan^2x+5tanx+2}dx\)
We substitute :
\(u=tanx,du=sec^2xdx\\ I=\int\dfrac{2}{2u^2+5u+2}du\\ =\int\dfrac{2}{2\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{8}}du\\ =\int\dfrac{1}{\left(u+\dfrac{5}{4}\right)^2-\dfrac{9}{16}}du\\ \)
Then,
\(t=u+\dfrac{5}{4}\\I=\int\dfrac{1}{t^2-\dfrac{9}{16}}dt\\ =\int\dfrac{\dfrac{2}{3}}{t-\dfrac{3}{4}}-\dfrac{\dfrac{2}{3}}{t+\dfrac{3}{4}}dt\)
Finally,
\(I=\dfrac{2}{3}ln\left(\left|\dfrac{t-\dfrac{3}{4}}{t+\dfrac{3}{4}}\right|\right)+C=\dfrac{2}{3}ln\left(\left|\dfrac{tanx+\dfrac{1}{2}}{tanx+2}\right|\right)+C\)
Bài 3.9:
a)
\(\int ^{1}_{0}(y^3+3y^2-2)dy=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{y^4}{4}+y^3-2y \right )=\frac{-3}{4}\)
b) \(\int ^{4}_{1}\left (t+\frac{1}{\sqrt{t}}-\frac{1}{t^2}\right)dt=\left.\begin{matrix} 4\\ 1\end{matrix}\right|\left ( \frac{t^2}{2}+2\sqrt{t}+\frac{1}{t} \right )=\frac{35}{4}\)
d) Ta có:
\(\int ^{1}_{0}(3^s-2^s)^2ds=\int ^{1}_{0}(9^s+4^s-2.6^s)ds=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{9^s}{\ln 9}+\frac{4^s}{\ln 4}-\frac{2.6^s}{\ln 6} \right )\)
\(=\frac{8}{\ln 9}+\frac{3}{\ln 4}-\frac{10}{\ln 6}\)
h)
Ta có \(\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx=\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{\sin^2x+\cos^2x+2\sin x\cos x}}dx\)
\(=\int ^{\frac{5\pi}{4}}_{\pi}\frac{-d(\sin x+\cos x)}{|\sin x+\cos x|}=\int ^{\frac{5\pi}{4}}_{\pi}\frac{d(\sin x+\cos x)}{\sin x+\cos x}=\left.\begin{matrix} \frac{5\pi}{4}\\ \pi\end{matrix}\right|\ln |\sin x+\cos x|=\ln (\sqrt{2})\)
Bài 3.10:
a)
Đặt \(t=1-x\) thì:
\(\int ^{2}_{1}x(1-x)^5dx=\int ^{-1}_{0}t^5(1-t)d(1-t)=\int ^{0}_{-1}t^5(1-t)dt\)
\(=\left.\begin{matrix} 0\\ -1\end{matrix}\right|\left ( \frac{t^6}{6}-\frac{t^7}{7} \right )=\frac{-13}{42}\)
b) Đặt \(\sqrt{e^x-1}=t\) \(\Rightarrow x=\ln (t^2+1)\)
Khi đó
\(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=\int ^{1}_{0}td(\ln (t^2+1))=\int ^{1}_{0}t.\frac{2t}{t^2+1}dt\)
\(=\int ^{1}_{0}\frac{2t^2}{t^2+1}dt=\int ^{1}_{0}2dt-\int ^{1}_{0}\frac{2}{t^2+1}dt=\left.\begin{matrix} 1\\ 0\end{matrix}\right|2t-\int ^{1}_{0}\frac{2dt}{t^2+1}=2-\int ^{1}_{0}\frac{2dt}{t^2+1}\)
Với \(\int ^{1}_{0}\frac{2dt}{t^2+1}\), đặt \(t=\tan m\)
\(\Rightarrow \int ^{1}_{0}\frac{2dt}{t^2+1}=\int ^{\frac{\pi}{4}}_{0}\frac{2d(\tan m)}{\tan ^2m+1}=\int ^{\frac{\pi}{4}}_{0}2\cos ^2md(\tan m)\)
\(=\int ^{\frac{\pi}{4}}_{0}2dm=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|2m=\frac{\pi}{2}\)
Do đó \(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=2-\frac{\pi}{2}\)
