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a) \(2^{x+4}+2^{x+2}=5^{x+1}+3\cdot5^x\)
\(\Rightarrow2^x+2^4+2x^x+2^2=5^x\cdot x+3\cdot5^x\)
\(\Leftrightarrow2^x+16+2^x\cdot4=5\cdot5^x+3\cdot5^x\)
\(\Leftrightarrow16\cdot2^x+4\cdot2^x=8\cdot5^x\)
\(\Leftrightarrow20\cdot2^x=8\cdot5^x\)
\(\Leftrightarrow20\cdot\left(\dfrac{2}{5}\right)^x=8\)
\(\Leftrightarrow\left(\dfrac{2}{5}\right)^x=\dfrac{2}{5}\)
\(\Leftrightarrow\left(\dfrac{2}{5}\right)^x=\left(\dfrac{2}{5}\right)^1\)
\(\Rightarrow x=1\)
\(y'=x^2-\left(3m+2\right)x+2m^2+3m+1\)
\(\Delta=\left(3m+2\right)^2-4\left(2m^2+3m+1\right)=m^2\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{3m+2+m}{2}=2m+1\\x_2=\frac{3m+2-m}{2}=m+1\end{matrix}\right.\)
Để hàm số có cực đại, cực tiểu \(\Rightarrow x_1\ne x_2\Rightarrow m\ne0\)
- Nếu \(m>0\Rightarrow2m+1>m+1\Rightarrow\left\{{}\begin{matrix}x_{CĐ}=m+1\\x_{CT}=2m+1\end{matrix}\right.\)
\(\Rightarrow3\left(m+1\right)^2=4\left(2m+1\right)\) \(\Rightarrow3m^2-2m-1=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-\frac{1}{3}< 0\left(l\right)\end{matrix}\right.\)
- Nếu \(m< 0\Rightarrow m+1>2m+1\Rightarrow\left\{{}\begin{matrix}x_{CĐ}=2m+1\\x_{CT}=m+1\end{matrix}\right.\)
\(\Rightarrow3\left(2m+1\right)^2=4\left(m+1\right)\Rightarrow12m^2+8m-1=0\)
\(\Rightarrow\left[{}\begin{matrix}m=\frac{-2+\sqrt{7}}{6}>0\left(l\right)\\m=\frac{-2-\sqrt{7}}{6}\end{matrix}\right.\) \(\Rightarrow\sum m=\frac{4-\sqrt{7}}{6}\)
1)
Ta có \(P_1=\int \frac{\cos xdx}{2\sin x-7}=\int \frac{d(\sin x)}{3\sin x-7}\)
Đặt \(\sin x=t\Rightarrow P_1=\int \frac{dt}{3t-7}=\frac{1}{3}\int \frac{d(3t-7)}{3t-7}=\frac{1}{3}\ln |3t-7|+c\)
\(=\frac{1}{3}\ln |3\sin x-7|+c\)
2)
\(P_2=\int \sin xe^{2\cos x+3}dx\)
Đặt \(\cos x=t\)
\(P_2=-\int e^{2\cos x+3}d(\cos x)=-\int e^{2t+3}dt\)
\(=-\frac{1}{2}\int e^{2t+3}d(2t+3)=\frac{-1}{2}e^{2t+3}+c\)
\(=\frac{-e^{2\cos x+3}}{2}+c\)
3)
\(P_3=\int \frac{\sin x+x\cos x}{(x\sin x)^2}dx\)
Để ý rằng \((x\sin x)'=x'\sin x+x(\sin x)'=\sin x+x\cos x\)
Do đó: \(d(x\sin x)=(x\sin x)'dx=(\sin x+x\cos x)dx\)
Suy ra \(P_3=\int \frac{d(x\sin x)}{(x\sin x)^2}\)
Đặt \(x\sin x=t\Rightarrow P_3=\int \frac{dt}{t^2}=\frac{-1}{t}+c=\frac{-1}{x\sin x}+c\)





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