Bài 1 Phân tích đa thức thành nhân tử a)5x ^2+20xy+20y^2-3x-6y b)(x^2+2)^2+12 Bài 2 Tìm x biết a) 4(x+1)^3+2(x+2)^2+1999=(x^2+2)(4x+14) b) 5(x^2-10x+25)-2(x-5)=0 giúp mình với ạ.Hứa vote 5 sao

Đáp án:

$\begin{array}{l}
B1)\\
a)5{x^2} + 20xy + 20{y^2} - 3x - 6y\\
 = 5.\left( {{x^2} + 4xy + 4{y^2}} \right) - 3\left( {x + 2y} \right)\\
 = 5.{\left( {x + 2y} \right)^2} - 3\left( {x + 2y} \right)\\
 = \left( {x + 2y} \right)\left( {5.\left( {x + 2y} \right) - 3} \right)\\
 = \left( {x + 2y} \right).\left( {5x + 10y - 3} \right)\\
b){\left( {{x^2} + 2} \right)^2} + 12\\
 = {x^4} + 4{x^2} + 4 + 12\\
 = {x^4} + 4{x^2} + 16\\
 = {x^4} + 8{x^2} + 16 - 4{x^2}\\
 = {\left( {{x^2} + 4} \right)^2} - {\left( {2x} \right)^2}\\
 = \left( {{x^2} + 4 - 2x} \right)\left( {{x^2} + 4 + 2x} \right)\\
B2)\\
a)4{\left( {x + 1} \right)^3} + 2{\left( {x + 2} \right)^2} + 1999\\
 = \left( {{x^2} + 2} \right)\left( {4x + 14} \right)\\
 \Leftrightarrow 4\left( {{x^3} + 3{x^2} + 3x + 1} \right) + 2\left( {{x^2} + 4x + 4} \right)\\
 + 1999 = 4{x^3} + 14{x^2} + 8x + 28\\
 \Leftrightarrow 4{x^3} + 12{x^2} + 12x + 4 + 2{x^2} + 8x + 8\\
 + 1999 = 4{x^3} + 14{x^2} + 8x + 28\\
 \Leftrightarrow 12x + 4 + 8 + 1999 = 28\\
 \Leftrightarrow 12x =  - 1984\\
 \Leftrightarrow x =  - \dfrac{{496}}{3}\\
\text{Vậy}\,x =  - \dfrac{{496}}{3}\\
b)5\left( {{x^2} - 10x + 25} \right) - 2\left( {x - 5} \right) = 0\\
 \Leftrightarrow 5{\left( {x - 5} \right)^2} - 2\left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {5\left( {x - 5} \right) - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {5x - 25 - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
5x - 27 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{{27}}{5}
\end{array} \right.\\
\text{Vậy}\,x = 5;x = \dfrac{{27}}{5}
\end{array}$