Bài 1: tìm X, biêt: a) (x +2)^2 + (x-3)^2 = 2x (x+7) b) (x+3).(x^2 – 3x +9) = x(x^2 +4) -1 c) (x+1)^3 +( x-1)^3 =2×3 d) (2x +1)^2 =16 e) x^2 -10x = 11

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{\left( {x + 2} \right)^2} + {\left( {x – 3} \right)^2} = 2x\left( {x + 7} \right)\\
 \Leftrightarrow \left( {{x^2} + 4x + 4} \right) + \left( {{x^2} – 6x + 9} \right) = 2{x^2} + 14x\\
 \Leftrightarrow 2{x^2} – 2x + 13 = 2{x^2} + 14x\\
 \Leftrightarrow 16x = 13\\
 \Leftrightarrow x = \dfrac{{13}}{{16}}\\
b,\\
\left( {x + 3} \right)\left( {{x^2} – 3x + 9} \right) = x\left( {{x^2} + 4} \right) – 1\\
 \Leftrightarrow {x^3} + {3^3} = {x^3} + 4x – 1\\
 \Leftrightarrow {x^3} + 27 = {x^3} + 4x – 1\\
 \Leftrightarrow 4x = 28\\
 \Leftrightarrow x = 7\\
c,\\
{\left( {x + 1} \right)^3} + {\left( {x – 1} \right)^3} = 2{x^3}\\
 \Leftrightarrow \left( {{x^3} + 3{x^2} + 3x + 1} \right) + \left( {{x^3} – 3{x^2} + 3x – 1} \right) = 2{x^3}\\
 \Leftrightarrow 2{x^3} + 6x = 2{x^3}\\
 \Leftrightarrow 6x = 0\\
 \Leftrightarrow x = 0\\
d,\\
{\left( {2x + 1} \right)^2} = 16\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 4\\
2x + 1 =  – 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  – \dfrac{5}{2}
\end{array} \right.\\
e,\\
{x^2} – 10x = 11\\
 \Leftrightarrow {x^2} – 10x – 11 = 0\\
 \Leftrightarrow \left( {{x^2} – 11x} \right) + \left( {x – 11} \right) = 0\\
 \Leftrightarrow x\left( {x – 11} \right) + \left( {x – 11} \right) = 0\\
 \Leftrightarrow \left( {x – 11} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x – 11 = 0\\
x + 1 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 11\\
x =  – 1
\end{array} \right.
\end{array}\)