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Đặt \(z=x+yi\)
\(\dfrac{\left(x+yi\right)\left(1+i\right)}{2}+\left(x-yi\right)\left(5+2i\right)=31-17i\)
\(\Leftrightarrow x-y+\left(x+y\right)i+10x+4xi-10yi+4y=31-17i\)
\(\Leftrightarrow\left(11x+3y\right)+\left(5x-9y\right)i=62-34i\)
\(\Leftrightarrow\left\{{}\begin{matrix}11x+3y=62\\5x-9y=-34\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Người ta có cho hàm f(x) là gì hay tích phân của hàm f(x) bằng bao nhiêu ko ?
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\(\int\limits^2_1\dfrac{2x+1}{x^2+x}dx=\int\limits^2_1\dfrac{d\left(x^2+x\right)}{x^2+x}=ln\left(x^2+x\right)|^2_1=ln6-ln2=ln3\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\b=3\end{matrix}\right.\) \(\Rightarrow a^2+b^2=9\)
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Có: \(I=\int\limits^{ln3}_0\sqrt{e^x-1}e^xdx\)
Đặt \(t=\sqrt{e^x-1}\Rightarrow t^2=e^x-1\)
\(\Rightarrow2tdt=e^xdx\)
\(\Rightarrow I=\int\limits^{\sqrt{2}}_02t^2dt\) \(\Rightarrow I=\dfrac{2}{3}t^3|^{\sqrt{2}}_0=\dfrac{4}{3}\sqrt{2}\)
Vậy a=0, b=\(\dfrac{4}{3}\) \(\Rightarrow a+b=0+\dfrac{4}{3}=\dfrac{4}{3}\)
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=100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000100000000000000000000000200000000000000000000122343534654654345646
nha bn !!
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