Giải thích các bước giải:
$4(x+5)(x+6)(x+10)(x+12)-3x^2$
$=4(x+5)(x+12)(x+6)(x+10)-3x^2$
$=4(x^2+17x+60)(x^2+16x+60)-3x^2$ $(1)$
$\text{Đặt: $t=x^2+16x+60$}$
$(1)$ $=4(t+x)t-3x^2$
$=4t^2+4xt-3x^2$
$=4t^2+4xt+x^2-4x^2$
$=(2t+x)^2-4x^2$
$=(2t+x-2x)(2t+x+2x)$
$=(2t-x)(2t+3x)$ $(2)$
$\text{Thay $x^2+16x+60$ vào lại ta được:}$
$(2)$ $=[2(x^2+16x+60)-x][2(x^2+16x+60)+3x]$
$=(2x^2+32x+120-x)(2x^2+32x+120+3x)$
$=(2x^2+31x+120)(2x^2+35x+120)$
$=(2x^2+16x+15x+120)\dfrac{1}{8}.8(2x^2+35x+120)$
$=\dfrac{1}{8}.[2x(x+8)+15(x+8)].(16x^2+280x+960)$
$=\dfrac{1}{8}.(x+8)(2x+15).(16x^2+2.4x.35+35^2-265)$
$=\dfrac{1}{8}.(x+8)(2x+15).[(4x+35)^2-265]$
$=\dfrac{1}{8}.(x+8)(2x+15).(4x+35-\sqrt{265})(4x+35+\sqrt{265})$
Chúc bạn học tốt !!!
