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Giống bài trước, \(x=3+2\sqrt{2}\) là nghiệm
\(\Rightarrow y=\dfrac{mx+1}{x-m}\Rightarrow y'=\dfrac{-m^2-1}{\left(x-m\right)^2}\) nghịch biến trên miền xác định
\(\Rightarrow\max\limits_{\left[1;2\right]}y=y\left(1\right)=\dfrac{m+1}{1-m}=-2\Rightarrow m\)
Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)
A. √3+1/2 B. √3−1/2 C. 1−√3/2 D. 0
Từ GT ta lấy tích phân 2 vế cận từ 0 đến 1 ; sẽ được :
\(\int\limits^1_0f\left(x+1\right)dx+\int\limits^1_03f\left(3x+2\right)dx-\int\limits^1_04f\left(4x+1\right)dx-\int\limits^1_0f\left(2^x\right)dx=\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}\left(1\right)\)
\(\int\limits^1_0\dfrac{3dx}{\sqrt{x+1}+\sqrt{x+2}}=\int\limits^1_03\left(\sqrt{x+2}-\sqrt{x+1}\right)dx\) =
\(2\left[\left(x+2\right)\sqrt{x+2}-\left(x+1\right)\sqrt{x+1}\right]\dfrac{1}{0}\) = \(2+6\sqrt{3}-8\sqrt{2}\left(2\right)\)
Dễ thấy : \(\int\limits^1_0f\left(x+1\right)dx=\int\limits^2_1f\left(t\right)dt=\int\limits^2_1f\left(x\right)dx\)
\(\int\limits^1_03f\left(3x+2\right)dx=\int\limits^5_2f\left(t\right)dt=\int\limits^5_2f\left(x\right)dx\) (3)
\(\int\limits^1_04f\left(4x+1\right)=\int\limits^5_1f\left(t\right)dt=\int\limits^5_1f\left(x\right)dx\left(4\right)\)
\(\int\limits^1_0f\left(2^x\right)dx=\int\limits^2_1\dfrac{f\left(t\right)dt}{tln2}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(t\right)dt}{t}=\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}\) (5)
Thay (2) ; (3) ; (4) ; (5) vào (1) ta được :
\(\int\limits^2_1f\left(x\right)dx+\int\limits^5_2f\left(x\right)dx-\int\limits^5_1f\left(x\right)dx-\dfrac{1}{ln2}.\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=2+6\sqrt{3}-8\sqrt{2}\)
\(\Leftrightarrow\int\limits^2_1\dfrac{f\left(x\right)dx}{x}=\left(2+6\sqrt{3}-8\sqrt{2}\right)ln2\)
a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)
\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)
b: \(y=\left(3x+1\right)^{\Omega}\)
=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)
=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)
c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)
\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)
\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)
d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)
\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)
\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)
\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)
e: \(y=3^{x^2}\)
=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)
f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)
h: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)
b) \(y=\left(3x+1\right)^{\pi}\)
\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)
c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)
d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)
e) \(y=3^{x^2}\)
\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)
f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)
Các bài còn lại bạn tự làm nhé!
Phương trình d dạng tham số: \(\left\{{}\begin{matrix}x=1+2t\\y=1+2t\\z=-1+t\end{matrix}\right.\)
Gọi \(M\left(1+2t;1+2t;-1+t\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AM}=\left(2t-5;2t+1;t-1\right)\\\overrightarrow{BM}=\left(2t+1;2t+1;t+5\right)\end{matrix}\right.\)
\(\Rightarrow P=\sqrt{\left(2t-5\right)^2+\left(2t+1\right)^2+\left(t-1\right)^2}+\sqrt{\left(2t+1\right)^2+\left(2t+1\right)^2+\left(t+5\right)^2}\)
\(=\sqrt{9t^2-18t+27}+\sqrt{9t^2+18t+27}\)
\(=\sqrt{\left(3-3t\right)^2+18}+\sqrt{\left(3+3t\right)^2+18}\)
\(\ge\sqrt{\left(3-3t+3+3t\right)^2+4.18}=6\sqrt{3}\)