\(y'=\left(3x^2\right)'-\left(4x\right)'+9'\)

\(y'=6x-4\Rightarrow y'\left(1\right)=6.1-4=2\)

Tham khảo:

Câu 5 ạ - Hoc24

\(S=1.3^0+2.3^1+3.3^2+...+11.3^{10}\)

\(3S=1.3^1+2.3^2+...+11.3^{11}\)

\(\Rightarrow S-3S=1+3^1+3^2+...+3^{10}-11.3^{11}\)

\(\Rightarrow-2S=1.\dfrac{3^{11}-1}{3-1}-11.3^{11}\)

\(\Rightarrow-2S=\dfrac{1}{2}.3^{11}-\dfrac{1}{2}-11.3^{11}\)

\(\Rightarrow-2S=-\dfrac{21.3^{11}+1}{2}\)

\(\Rightarrow S=\dfrac{1}{4}+\dfrac{21.3^{11}}{4}\)

\(f'\left(x\right)=\dfrac{1}{2\sqrt{2+x}}-\dfrac{1}{2\sqrt{7-x}}+\dfrac{5-2x}{2\sqrt{\left(2+x\right)\left(7-x\right)}}\)

\(f'\left(x\right)\) không xác định khi \(\left[{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

\(f'\left(x\right)=0\Rightarrow\sqrt{7-x}-\sqrt{2+x}+5-2x=0\)

\(\Rightarrow x=\dfrac{5}{2}\)

tham khảo

Với k \(\in\)N* ; ta có : \(kC_n^k=k.\dfrac{n!}{\left(n-k\right)!k!}=\dfrac{n!}{\left(n-k\right)!\left(k-1\right)!}=\dfrac{n\left(n-1\right)!}{\left[n-1-\left(k-1\right)\right]!\left(k-1\right)!}=nC_{n-1}^{k-1}\)

Khi đó : \(C_n^1+2C_n^2+...+nC^n_n\)  = \(\Sigma^n_{k=1}nC^{k-1}_{n-1}\)  

= \(n\left(C_{n-1}^0+C_{n-1}^1+...+C_{n-1}^{n-1}\right)\)  \(=n.\left(1+1\right)^{n-1}=n.2^{n-1}\) ( đpcm )

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[]{2x+1}-\sqrt[9]{9x+1}}{\sqrt[]{4x+1}-1}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{2x+1}-1\right)-\left(\sqrt[3]{9x+1}-1\right)}{\sqrt[]{4x+1}-1}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt[]{2x+1}+1}-\dfrac{9x}{\sqrt[3]{\left(9x+1\right)^2}+\sqrt[3]{9x+1}+1}}{\dfrac{4x}{\sqrt[]{4x+1}+1}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2}{\sqrt[]{2x+1}+1}-\dfrac{9}{\sqrt[3]{\left(9x+1\right)^2}+\sqrt[3]{9x+1}+1}}{\dfrac{4}{\sqrt[]{4x+1}+1}}=\dfrac{1-3}{2}=-1\)

\(y'=\dfrac{13}{\left(x+3\right)^2}\) \(\Rightarrow\left\{{}\begin{matrix}y'\left(10\right)=\dfrac{1}{13}\\y\left(10\right)=3\end{matrix}\right.\)

Phương trình tiếp tuyến:

\(y=\dfrac{1}{13}\left(x-10\right)+3\Leftrightarrow y=\dfrac{1}{13}x+\dfrac{29}{13}\)

1. \(y'=-8x^3+8x-3\)

2. \(y'=3x^2-6x+1\)

3. \(y'=\dfrac{5}{2}x^4+4x^3-3x^2-3x+4\)

4. \(y'=-2x^3+2x-\dfrac{1}{3}\)

1. 

2. 

3. 

4. 

đây nha

1.

\(y'=-8x^3+8x-3\)

2.

\(y'=3x^2-6x+1\)

3.

\(y'=\dfrac{5}{2}x^4+4x^3-3x^2-3x+4\)

4.

\(y'=-\dfrac{1}{3}+2x-2x^3\)

Em cảm ơn ạ <3