1/ x^2 = 4x+5 tìm X 2/ B= 2.(3^2 +1)(3^4+1)(3^16+1)(3^32+1) 3/ tìm a để đa thức 2x^3 – 3x^2+x+ chia hết cho x+2 4/ tìm các giá trị nguyên củ

Giải thích các bước giải:

\(\begin{array}{l}
1){x^2} = 4x + 5\\
 \Leftrightarrow {x^2} – 4x – 5 = 0\\
 \Leftrightarrow {x^2} – 5x + x – 5 = 0\\
 \Leftrightarrow x\left( {x – 5} \right) + \left( {x – 5} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x – 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 1\\
x = 5
\end{array} \right.\\
2)B = \dfrac{1}{4}\left( {{3^2} – 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)\\
 = \dfrac{1}{4}\left( {{3^4} – 1} \right)\left( {{3^4} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)\\
 = \dfrac{1}{4}.\dfrac{1}{{{3^8} + 1}}\left( {{3^8} – 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)\\
 = \dfrac{1}{{4\left( {{3^8} + 1} \right)}}\left( {{3^{16}} – 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)\\
 = \dfrac{1}{{4\left( {{3^8} + 1} \right)}}\left( {{3^{64}} – 1} \right)\\
3)2{x^3} + 4{x^2} – 7{x^2} – 14x + 15x + 30 – 30 + a\\
 = 2{x^2}\left( {x + 2} \right) – 7x\left( {x + 2} \right) + 15\left( {x + 2} \right) + a – 30\\
 = \left( {x + 2} \right)\left( {2{x^2} – 7x + 15} \right) + a – 30\\
gt \Rightarrow a – 20 = 0 \Leftrightarrow a = 30\\
4)\dfrac{3}{{x – 2}} \in Z \Rightarrow \left( {x – 2} \right) \in \left\{ { – 1;1; – 3;3} \right\}\\
 \Rightarrow x \in \left\{ {1;3; – 1;5} \right\}
\end{array}\)