1 câu trả lời
$\left(x²+3x+1\right)²+\left(3x+1\right)²-2\left(x²+3x+1\right)\left(3x-1\right)$
$=x^4+6x^3+11x^2+6x+1+\left(3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)$
$=x^4+6x^3+11x^2+6x+1+9x^2+6x+1-2\left(x^2+3x+1\right)\left(3x-1\right)$
$=x^4+6x^3+11x^2+6x+1+9x^2+6x+1-6x^3-16x^2+2$
$=x^4+6x^3-6x^3+11x^2+9x^2-16x^2+6x+6x+1+1+2$
$=x^4+11x^2+9x^2-16x^2+6x+6x+1+1+2$
$=x^4+11x^2+9x^2-16x^2+6x+6x+4$
$=x^4+4x^2+6x+6x+4$
$=x^4+4x^2+12x+4$
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