Bài 9:

Đặt f(x) = \(2x^3+ax+b\)

Vì f(x) = \(2x^3+ax+b\) chia cho x + 1 dư 6 và chia cho x - 2 dư 21 nên ta có:

\(\hept{\begin{cases}f\left(-1\right)=2\times\left(-1\right)^3-a+b=6\\f\left(2\right)=2\times2^3+2a+b=21\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-2-a+b=6\\16+2a+b=21\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-a+b=8\\2a+b=5\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3a=3\\b=5-2a\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=-1\\b=7\end{cases}}\)

Vậy a = -1, b = 7

nani ~
 

What???

P=\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

bài này phải ko

Ngố ngây ngô là mình

Mk làm bài 2 nha ><

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#Yiin - girl ><

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#Yiin - girl ><