câu 1 (5+3)(5^2+3^2)(5^4+3^4)……(5^64+3^64)+5^128-3^128/2 câu 2 3x(x+2)-2(x^2-5x)-x(x+10)=12 . giúp mình nha mọi người

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
A = \left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right) + \dfrac{{{5^{128}} – {3^{128}}}}{2}\\
B = \left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = 2.\left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {5 – 3} \right).\left( {5 + 3} \right)\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {{5^2} – {3^2}} \right).\left( {{5^2} + {3^2}} \right)\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = \left( {{5^4} – {3^4}} \right).\left( {{5^4} + {3^4}} \right)…..\left( {{5^{64}} + {3^{64}}} \right)\\
….\\
 \Leftrightarrow 2B = \left( {{5^{64}} – {3^{64}}} \right)\left( {{5^{64}} + {3^{64}}} \right)\\
 \Leftrightarrow 2B = {5^{128}} – {3^{128}}\\
 \Leftrightarrow B = \dfrac{{{5^{128}} – {3^{128}}}}{2}\\
A = B + \dfrac{{{5^{128}} – {3^{128}}}}{2} = {5^{128}} – {3^{128}}\\
2,\\
3x\left( {x + 2} \right) – 2\left( {{x^2} – 5x} \right) – x\left( {x + 10} \right) = 12\\
 \Leftrightarrow \left( {3{x^2} + 6x} \right) – \left( {2{x^2} – 10x} \right) – \left( {{x^2} + 10x} \right) = 12\\
 \Leftrightarrow 3{x^2} + 6x – 2{x^2} + 10x – {x^2} – 10x = 12\\
 \Leftrightarrow 6x = 12\\
 \Leftrightarrow x = 2
\end{array}\)