Giúp mk với 1, Phân tích đa thức sau thành nhân tử: a, x^2 – 2x – 7 b, 4x^2 + 4x – 2 c, 9x^2 + 12x + 1 d, 4x^2 + 6xy – 3 2, Phân tích đa thức sau thàn

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
{x^2} – 2x – 7 = \left( {{x^2} – 2x + 1} \right) – 8 = {\left( {x – 1} \right)^2} – {\left( {2\sqrt 2 } \right)^2} = \left( {x – 1 – 2\sqrt 2 } \right)\left( {x – 1 + 2\sqrt 2 } \right)\\
b,\\
4{x^2} + 4x – 2 = \left( {4{x^2} + 4x + 1} \right) – 3 = {\left( {2x + 1} \right)^2} – {\sqrt 3 ^2} = \left( {2x + 1 – \sqrt 3 } \right)\left( {2x + 1 + \sqrt 3 } \right)\\
c,\\
9{x^2} + 12x + 1 = {\left( {3x} \right)^2} + 2.3x.2 + 4 – 3 = {\left( {3x + 2} \right)^2} – {\sqrt 3 ^2} = \left( {3x + 2 – \sqrt 3 } \right)\left( {3x + 2 + \sqrt 3 } \right)\\
2,\\
a,\\
{x^2} – 5x + 6 = \left( {{x^2} – 2x} \right) – \left( {3x – 6} \right) = x\left( {x – 2} \right) – 3\left( {x – 2} \right) = \left( {x – 2} \right)\left( {x – 3} \right)\\
b,\\
{x^2} – 5x – 6 = \left( {{x^2} – 6x} \right) + \left( {x – 6} \right) = x\left( {x – 6} \right) + \left( {x – 6} \right) = \left( {x – 6} \right)\left( {x + 1} \right)\\
c,\\
4{x^2} – 16x + 7 = \left( {4{x^2} – 2x} \right) – \left( {14x – 7} \right) = 2x\left( {2x – 1} \right) – 7\left( {2x – 1} \right) = \left( {2x – 1} \right)\left( {2x – 7} \right)\\
d,\\
 – 6{x^2} + 7x – 2 = \left( { – 6{x^2} + 3x} \right) + \left( {4x – 2} \right) =  – 3x\left( {x – 2} \right) + 2\left( {x – 2} \right) = \left( {x – 2} \right)\left( { – 3x + 2} \right)\\
e,\\
3{x^2} + x – 2 = \left( {3{x^2} + 3x} \right) – \left( {2x + 2} \right) = 3x\left( {x + 1} \right) – 2\left( {x + 1} \right) = \left( {x + 1} \right)\left( {3x – 2} \right)\\
f,\\
3{x^2} + 10x + 3 = \left( {3{x^2} + 9x} \right) + \left( {x + 3} \right) = 3x\left( {x + 3} \right) + \left( {x + 3} \right) = \left( {x + 3} \right)\left( {3x + 1} \right)\\
g,\\
{x^2} – x – 12 = \left( {{x^2} – 4x} \right) + \left( {3x – 12} \right) = x\left( {x – 4} \right) + 3\left( {x – 4} \right) = \left( {x – 4} \right)\left( {x + 3} \right)
\end{array}\)