Xet \(m\ne-3\)

\(=\lim\limits_{x\rightarrow-\infty}x\left(\sqrt[3]{1}+\sqrt{4}+m\right)=x\left(3+m\right)\)

\(=\left[{}\begin{matrix}-\infty\left(m>-3\right)\\+\infty\left(m< -3\right)\end{matrix}\right.\)

Xet \(m=-3\)

\(=\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+2x^2+1}-x-2x-\sqrt{4x^2+2x+3}\right)\)

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

\(=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)

Minh thu thoi chu minh ko chac a!

\(z=x+yi\left(x,y\in R\right)\)

\(\left|x+yi-x+yi+1-i\right|=\sqrt{5}=\left|1+\left(2y-1\right)i\right|\)

\(\Leftrightarrow1+\left(2y-1\right)^2=5\Leftrightarrow2y-1=\pm2\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vi \(\left(2-z\right)\left(i+\overline{z}\right)\) la so thuan ao

\(\Rightarrow\left(2-x-yi\right)\left(i+x-yi\right)=2i+2x-2yi-xi-x^2+yxi+y-xyi-y^2=\left(2-2y-x\right)i+2x+y-\left(x^2+y^2\right)\) la so thuan ao

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y-\left(x^2+y^2\right)=0\\2-2y-x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+y^2+y=0\\x\ne2-2y\end{matrix}\right.\)

Voi \(y=\dfrac{3}{2}\Rightarrow x^2-2x+\dfrac{9}{4}+\dfrac{3}{2}=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Voi \(y=-\dfrac{1}{2}\Rightarrow x^2-2x+\dfrac{1}{4}-\dfrac{1}{2}=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2+\sqrt{5}}{2}\\x=\dfrac{2-\sqrt{5}}{2}\end{matrix}\right.\)

=> 4 so phuc z thoa man

\(=\lim\limits_{x\rightarrow2}\dfrac{\sqrt[3]{9}-\sqrt{5}}{0}=-\infty\)

P/s: Bạn coi lại đề bài ạ, mình cá là bài này phải cho thành dạng 0/0 chứ ko đơn thuần là thay vô xong ngay được!

\(P_{phat-quang}=N_{phat-quang}.\dfrac{hc}{\lambda_{phat-quang}}\)

\(P_{kich-thich}=N_{kich-thich}.\dfrac{hc}{\lambda_{kich-thich}}\)

\(\Rightarrow\dfrac{N_{pq}}{N_{kt}}=\dfrac{\dfrac{P_{pq}}{hc}.\lambda_{pq}}{\dfrac{P_{kt}}{hc}.\lambda_{kt}}=\dfrac{P_{pq}}{P_{kt}}.\dfrac{\lambda_{pq}}{\lambda_{kt}}=0,1.\dfrac{0,5}{0,3}=\dfrac{1}{6}\)

\(f\left(0\right)=-1\Rightarrow f'\left(0\right)+2=0\Leftrightarrow f'\left(0\right)=-2\)

\(\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\dfrac{f'\left(x\right)-x.e^{3x}}{2}dx=\dfrac{1}{2}\int\limits^1_0f'\left(x\right)dx-\dfrac{1}{2}\int\limits^1_0x.e^{3x}dx=\dfrac{1}{2}f\left(x\right)|^1_0-\dfrac{1}{2}\int\limits^1_0xe^{3x}dx\)

\(I_1=\int xe^{3x}dx\)

\(\left\{{}\begin{matrix}u=x\\dv=e^{3x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=dx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{1}{3}xe^{3x}-\dfrac{1}{3}\int e^{3x}dx=\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\)

\(\Rightarrow I=\dfrac{1}{2}f\left(1\right)-\dfrac{1}{2}f\left(0\right)-\dfrac{1}{2}\left(\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\right)|^1_0\)

Èo, tắc chỗ f(1) rồi, vậy đành phải biến đổi để tìm f(x) luôn vậy, hmm

Thử nhân 2 vế với \(e^{2x}\) xem nào:

\(e^{2x}f'\left(x\right)-2e^{2x}f\left(x\right)=x.e^{5x}\Leftrightarrow\left(e^{2x}.f\left(x\right)\right)'=x.e^{5x}\)

Lay nguyen ham 2 ve:

\(e^{2x}.f\left(x\right)=\int x.e^{5x}dx\)

\(\left\{{}\begin{matrix}x=u\\dv=e^{5x}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}dx=du\\v=\dfrac{1}{5}e^{5x}\end{matrix}\right.\)

\(\Rightarrow e^{2x}.f\left(x\right)=\int x.e^{5x}dx=\dfrac{1}{5}x.e^{5x}-\dfrac{1}{5}\int e^{5x}dx=\dfrac{1}{5}xe^{5x}-\dfrac{1}{25}e^{5x}+C\)

\(f\left(0\right)=-1\Leftrightarrow f\left(0\right)=-\dfrac{1}{25}+C=-1\Leftrightarrow C=-\dfrac{24}{25}\)

\(\Rightarrow f\left(x\right)=\dfrac{\dfrac{1}{5}xe^{5x}-\dfrac{1}{25}e^{5x}-\dfrac{24}{25}}{e^{2x}}\)

Vậy là xong rồi \(\Rightarrow f\left(1\right)=...\) , thay vô \(I=\dfrac{1}{2}f\left(1\right)-\dfrac{1}{2}.\left(-1\right)-\dfrac{1}{2}\left(\dfrac{1}{3}xe^{3x}-\dfrac{1}{9}e^{3x}\right)|^1_0\) là được nha :)

Ptdt đi qua M có dạng: \(y=y_0+k\left(x-x_0\right)\Leftrightarrow y=2+k\left(x-1\right)=kx-k+2\)

Vi dt do tiep xuc voi do thi

\(\Rightarrow\left\{{}\begin{matrix}x^3-2x^2+\left(m-1\right)x+2m=kx-k+2\left(1\right)\\3x^2-4x+m-1=k\left(2\right)\end{matrix}\right.\)

The (2) vo (1)

\(x^3-2x^2+\left(m-1\right)x+2m=\left(3x^2-4x+m-1\right)x-3x^2+4x-m+1+2\)

\(\Leftrightarrow2x^3-5x^2+4x+3=3m\) (3)

Xet \(f\left(x\right)=2x^3-5x^2+4x+3\Rightarrow f'\left(x\right)=6x^2-10x+4\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=1\end{matrix}\right.\)

Nhìn vô bbt, để (3) có 2 nghiệm pb (do có 2 tiếp tuyến) thì đường thẳng y=3m phải cắt đt tại 2 điểm\(\Rightarrow\left[{}\begin{matrix}3m=\dfrac{109}{27}\\3m=4\end{matrix}\right.\Rightarrow...\)

Cứ mỗi lần anh Lâm onl là ông đăng bài hỏi với tốc độ bàn thờ :v

a/ Hoành độ giao điểm của (C) với trục tung là \(x_0=0\)

\(y'=x^2-2x+2\)

\(\Rightarrow pttt:y-y_0=y'\left(x-x_0\right)\Leftrightarrow y=1+2x\)

b/ \(y'=x^2-2x+2\)

Goi \(M\left(x_0;y_0\right)\) la tiep diem \(\Rightarrow k=y'=x_0^2-2x_0+2\)

Vi tiep tuyen vuong goc voi \(y=-\dfrac{1}{5}x+2\)

\(\Rightarrow k.k'=-1\Leftrightarrow\left(x_0^2-2x_0+2\right).\left(-\dfrac{1}{5}\right)=-1\Leftrightarrow x_0^2-2x_0+2=5\)

\(\Leftrightarrow\left[{}\begin{matrix}x_0=3\\x_0=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y_0=\dfrac{3^3}{3}-3^2+2.3+1=7\\y_0=-\dfrac{1}{3}-1-2+1=-\dfrac{7}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=7+5\left(x-3\right)\\y=-\dfrac{7}{3}+5\left(x+1\right)\end{matrix}\right.\)

P/s: Check lại số hộ mình ạ!