Question

1) Cho 2 số dương thỏa a¹⁰ + b¹⁰ = a¹¹ + b¹¹ = a¹² + b¹². Tính P = a²⁰ + b²⁰

2) Cho \(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{a+c}=2012\). Tính A = \(\dfrac{b^2}{a+b}+\dfrac{c^2}{b+c}+\dfrac{a^2}{a+c}\)

3) Tìm a để x³ + 3ax² + 3a²x + a³ chia hết cho x² + 4x + 4

4) Cho a + b = a³ + b³ = -1. Tính (a - b)²⁰¹⁸

5) Cho a³ - 3ab² = 2 và b³ - 3a²b = -11. Tính a² + b²

Answer 1

Câu 1:

Theo bài ra ta có:

\(a^{12}+b^{12}=a^{12}+a^{11}b-a^{11}b-ab^{11}+ab^{11}+b^{12}\)

\(=a^{11}\left(a+b\right)-ab\left(a^{10}+b^{10}\right)+b^{11}\left(a+b\right)\)

\(=\left(a+b\right)\left(a^{11}+b^{11}\right)-ab\left(a^{10}+b^{10}\right)\)

\(=\left(a+b\right)\left(a^{12}+b^{12}\right)-ab\left(a^{12}+b^{12}\right)\)(gt cho rồi nhé)

\(=\left(a^{12}+b^{12}\right)\left(a+b-ab\right)\)

\(\Rightarrow a+b-ab=1\)

\(\Leftrightarrow a+b-ab-1=0\)

\(\Leftrightarrow a\left(1-b\right)-\left(1-b\right)=0\)

\(\Leftrightarrow\left(1-b\right)\left(a-1\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}b=1\\a=1\end{matrix}\right.\)

=> a^20 + b^20 = 2

:)) đừng ném đá nhá