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Tham khảo:

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9 tháng 4 2017

Ta có f'(x) = 2x, suy ra f'(1) = 2

và φ'(x) = 4 + . cos = 4 + . cos, suy ra φ'(1) = 4.

Vậy = = .

Tham khảo:

Giải sách bài tập Toán 11 | Giải sbt Toán 11

7 tháng 5 2016

Theo định nghĩa ta có :

\(f'\left(x\right)=\lim\limits_{\Delta x\rightarrow0}\frac{f\left(a+\right)-f\left(a\right)}{\Delta x}\)

         \(=\lim\limits_{\Delta x\rightarrow0}\frac{\left(a+\Delta x-1\right)\varphi\left(a+\Delta x\right)}{\Delta x}\) do (\(f\left(a\right)=0\))

          \(=\lim\limits_{\Delta x\rightarrow0}\varphi\left(a+\Delta x\right)\)

Khi \(\Delta x\rightarrow0\) thì \(a+\Delta x\rightarrow a\) và do \(\varphi\left(x\right)\) là hàm liên tục tại x = a nên có :

\(\lim\limits_{\Delta x\rightarrow0}\varphi\left(a+\Delta x\right)=\varphi\left(a\right)\)

Vậy \(f'\left(a\right)=\varphi\left(a\right)\)

AH
Akai Haruma
Giáo viên
12 tháng 7 2017

Câu a)

Đặt \(y=\sqrt{t}\Rightarrow I_1=\int ^{1}_{0}(y-1)^2\sqrt{y}dy=\int ^{1}_{0}(t^2-1)^2td(t^2)\)

\(\Leftrightarrow I_1=2\int^{1}_{0}(t^2-1)^2t^2dt=2\int ^{1}_{0}(t^6-2t^4+t^2)dt\)

\(=2\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{t^7}{7}-\frac{2t^5}{5}+\frac{t^3}{3} \right )=\frac{16}{105}\)

b) Đặt \(u=\sqrt[3]{z-1}\Rightarrow z=u^3+1\Rightarrow I_2=\int ^{1}_{0}[(u^3+1)^2+1]u^2d(u^3+1)\)

\(\Leftrightarrow I_2=3\int ^{1}_{0}[(u^3+1)^2+1]u^4du=3\int ^{1}_{0}(u^{10}+2u^7+2u^4)du\)

\(=3\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{x^{11}}{11}+\frac{x^8}{4}+\frac{2x^5}{5} \right )=\frac{489}{220}\)

AH
Akai Haruma
Giáo viên
12 tháng 7 2017

c) Ta có:

\(I_3=\int ^{e}_{1}\frac{\sqrt{4+5\ln x}}{x}dx=\int ^{e}_{1}\sqrt{4+5\ln x}d(\ln x)\)

Đặt \(\sqrt{4+5\ln x}=t\Rightarrow I_3=\int ^{3}_{2}td\left (\frac{t^2-4}{5}\right)=\frac{2}{5}\int ^{3}_{2}t^2dt=\frac{38}{15}\)

d)

Xét \(\int ^{\frac{\pi}{2}}_{0}\cos ^5xdx=\int ^{\frac{\pi}{2}}_{0}\cos ^4xd(\sin x)=\int ^{\frac{\pi}{2}}_{0}(1-\sin ^2x)^2d(\sin x)\)

\(=\int ^{1}_{0}(1-t^2)^2dt\)

Xét \(\int ^{\frac{\pi}{2}}_{0}\sin ^5xdx=-\int ^{\frac{\pi}{2}}_{0}\sin ^4xd(\cos x)=-\int ^{\frac{\pi}{2}}_{0}(1-\cos ^2x)^2d(\cos x)=\int ^{1}_{0}(1-t^2)^2dt\)

Do đó \(\int ^{\frac{\pi}{2}}_{0}(\cos ^5x-\sin ^5x)dx=0\)

e)

\(\int \cos ^3x\cos 3xdx=\int \cos 3x\left ( \frac{3\cos x+\cos 3x}{4} \right )dx=\frac{1}{4}\int \cos ^23xdx+\frac{3}{4}\int \cos x\cos 3xdx\)

\(=\frac{1}{8}\int (1+\cos 6x)dx+\frac{3}{8}\int (\cos 4x+\cos 2x)dx\)

\(=\frac{1}{8}\int (1+\cos 6x)dx+\frac{3}{8}\int (\cos 4x+\cos 2x)dx=\frac{x}{8}+\frac{\sin 6x}{48}+\frac{3\sin 4x}{32}+\frac{3\sin 2x}{16}\)

Suy ra \(\int ^{\pi}_{0}\cos ^3x\cos 3xdx=\frac{\pi}{8}\)

14 tháng 8 2023

a/

\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)

\(\Rightarrow x=12\)

 

 

14 tháng 8 2023

\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)

 

 

\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)

 

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)

\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)

\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)

Vậy x = 2023

18 tháng 6 2023

Ta có \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c=\dfrac{a+b+c}{2}\)

Suy ra

M = (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) + x2

    = x2 - ax - bx + ab + x2 - bx - cx + bc + x2 - ax - cx + ac + x2

    = 4x2 - 2ax - 2bx - 2cx + ab + bc + ac

    = (2x)2 - 2x(a + b + c) + ab + bc + ac

    = \(\left(2\cdot\dfrac{a+b+c}{2}\right)^2-\left(2\cdot\dfrac{a+b+c}{2}\right)\left(a+b+c\right)+ab+bc+ac\)

    = ab + bc + ac

AH
Akai Haruma
Giáo viên
13 tháng 7 2018

Lời giải:

PT \(\Leftrightarrow \frac{(x+4)-(x+2)}{(x+2)(x+4)}+\frac{(x+8)-(x+4)}{(x+4)(x+8)}+\frac{(x+14)-(x+8)}{(x+8)(x+14)}=\frac{x}{(x+2)(x+14)}\)

\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)

\(\Leftrightarrow \frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{(x+2)(x+14)}\)

\(\Leftrightarrow \frac{12}{(x+2)(x+14)}=\frac{x}{(x+2)(x+14)}\)

\(\Rightarrow x=12\) (thỏa mãn)

Vậy......