Đáp án:
$\begin{array}{l}
1.61\\
c)A = {\left( {{x^2} + x - 1} \right)^2} + 4{x^2} + 4x\\
= {\left( {{x^2} + x - 1} \right)^2} + 4\left( {{x^2} + x - 1} \right) + 4\\
Dat:{x^2} + x - 1 = a\\
\Rightarrow A = {a^2} + 4a + 4\\
= {\left( {a + 2} \right)^2}\\
= {\left( {{x^2} + x - 1 + 2} \right)^2}\\
= {\left( {{x^2} + x + 1} \right)^2}\\
1.64.\\
a){x^3} + 15{x^2} + 75x + 125 = 0\\
\Rightarrow {x^3} + 3.{x^2}.5 + 3.x{.5^2} + {5^3} = 0\\
\Rightarrow {\left( {x + 5} \right)^3} = 0\\
\Rightarrow x = - 5\\
b){\left( {x + 2} \right)^3} + {\left( {x - 3} \right)^3} = 0\\
\Rightarrow {\left( {x + 2} \right)^3} = - {\left( {x - 3} \right)^3}\\
\Rightarrow {\left( {x + 2} \right)^3} = {\left( {3 - x} \right)^3}\\
\Rightarrow x + 2 = 3 - x\\
\Rightarrow 2x = 1\\
\Rightarrow x = \frac{1}{2}\\
1.65\\
{x^2} + 102 = {y^2}\\
\Rightarrow {x^2} - {y^2} = 102\\
\Rightarrow \left( {x - y} \right)\left( {x + y} \right) = 102\\
= 1.102 = 2.51 = 3.34 = 4.28 = 7.16\\
Do:\left( {x - y} \right) + \left( {x + y} \right) = 2x \vdots 2
\end{array}$
=> tổng (x+y) và (x-y) phải là 1 số chẵn
$ \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + y = 4\\
x - y = 28
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = 28\\
x - y = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = - 4\\
x - y = - 28
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y = - 28\\
x - y = - 4
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 16;y = - 12\\
x = 16;y = 12\\
x = - 16;y = 12\\
x = - 16;y = - 12
\end{array} \right.$
