=> (x + 2)(x + 8)(x + 4)(x + 6) + 16 =0
=> (x2 + 10x + 16)(x2 + 10x + 24) + 16 = 0
nhân vào , rút gon ta đc :
x4 + 20x3 + 140x2 + 400x + 400 = 0
=> x4 + 100x2 + 400 + 20x2 + 400x + 40x2 =0
=> (x2 + 10x + 20)2 = 0
=> x2 + 10x + 20 = 0
Tính denta ra ta đc : x1 = \(\sqrt{5}-5\) ; x2 = \(-\sqrt{5}-5\)
ẩn phụ đi :v
( x + 2 )( x + 4 )( x + 6 )( x + 8 ) + 16 = 0
<=> [ ( x + 2 )( x + 8 ) ][ ( x + 4 )( x + 6 ) ] + 16 = 0
<=> ( x2 + 10x + 16 )( x2 + 10x + 24 ) + 16 = 0
Đặt t = x2 + 10x + 20
<=> ( t - 4 )( t + 4 ) + 16 = 0
<=> t2 - 16 + 16 = 0
<=> t2 = 0
<=> ( x2 + 10x + 20 )2 = 0
<=> x2 + 10x + 20 = 0
Δ = b2 - 4ac = 102 - 4.1.20 = 100 - 80 = 20
Δ > 0 nên phương trình có hai nghiệm phân biệt
\(x_1=\frac{-b+\sqrt{\text{Δ}}}{2a}=\frac{-10+\sqrt{20}}{2}=-5+\sqrt{5}\)
\(x_2=\frac{-b-\sqrt{\text{Δ}}}{2a}=\frac{-10-\sqrt{20}}{2}=-5-\sqrt{5}\)
Vậy \(x=-5\pm\sqrt{5}\)
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16=0\)
\(\Leftrightarrow\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16=0\)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)
Đặt \(x^2+10x+16=t\left(t\ge0\right)\)
\(t\left(t+8\right)+16=0\)
\(\Leftrightarrow t^2+8t+16=0\Leftrightarrow\left(t+4\right)^2=0\)
\(\Leftrightarrow t=-4\)hay \(x^2+10x+16=-4\)
\(\Leftrightarrow x^2+10x+20=0\)giải delta ta được :
\(x=-5\pm\sqrt{5}\)
lớp 8 mà , làm gì mà đen ta vậy mấy bạn
\(x^2+10x+20=0< =>x^2+10x+5^2-\sqrt{5}^2=0\)
\(< =>\left(x+5\right)^2-\sqrt{5}^2=0< =>\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)
\(< =>\orbr{\begin{cases}x=\sqrt{5}-5\\x=-\sqrt{5}-5\end{cases}}\)
