Question

[Thao Thanh]

giải giúp hết với

câu 1. giải phương trình

a) 4.(x+5).(x+6).(x+10).(x+12) = \(3x^2\)

b) (x+1).(x+2).(x+3).(x+6) = \(120x^2\)

c) (x+2).(x+3).(x+8).(x+12) = \(4x^2\)

d) (x-4).(x-5).(x-8).(x-10) = \(72x^2\)

e) (x+1).(x+3).(x+5).(x+15) = \(-16x^2\)

f) (\(2x^2\) - 3x + 1).(\(2x^2\) + 5x + 1) = \(9x^2\)

Answer 1

[Akai Haruma]

a.

$4(x+5)(x+6)(x+10)(x+12)=3x^2$

$4[(x+5)(x+12)][(x+6)(x+10)]=3x^2$

$4(x^2+17x+60)(x^2+16x+60)=3x^2$

Đặt $x^2+16x+60=a$ thì pt trở thành:

$4(a+x)a=3x^2$

$4a^2+4ax-3x^2=0$

$4a^2-2ax+6ax-3x^2=0$

$2a(2a-x)+3x(2a-x)=0$

$(2a-x)(2a+3x)=0$

Nếu $2a-x=0\Leftrightarrow 2(x^2+16x+60)-x=0$

$\Leftrightarrow 2x^2+31x+120=0\Rightarrow x=\frac{-15}{2}$ hoặc $x=-8$

Nếu $2a+3x=0\Leftrightarrow 2(x^2+16x+60)+3x=0$

$\Leftrightarrow 2x^2+35x+120=0\Rightarrow x=\frac{-35\pm \sqrt{265}}{4}$

Answer 2

[Akai Haruma]

b.

$(x+1)(x+2)(x+3)(x+6)=120x^2$

$[(x+1)(x+6)][(x+2)(x+3)]=120x^2$

$(x^2+7x+6)(x^2+5x+6)=120x^2$

Đặt $x^2+6=a$ thì pt trở thành:

$(a+7x)(a+5x)=120x^2$

$\Leftrightarrow a^2+12ax-85x^2=0$

$\Leftrightarrow a^2-5ax+17ax-85x^2=0$

$\Leftrightarrow a(a-5x)+17x(a-5x)=0$

$\Leftrightarrow (a-5x)(a+17x)=0$

Nếu $a-5x=0\Leftrightarrow x^2+6-5x=0$

$\Leftrightarrow (x-2)(x-3)=0\Rightarrow x=2$ hoặc $x=3$

Nếu $a+17x=0\Leftrightarrow x^2+17x+6=0$

$\Rightarrow x=\frac{-17\pm \sqrt{265}}{2}$

Vậy.........