\(\Leftrightarrow2\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=3,5\times2=7\)
\(\Leftrightarrow\left(8x+7\right)^2\left[2\left(4x+3\right)\right]\left[8\left(x+1\right)\right]=7\times8\)
Đặt t=8x+7.Pt trở thành:
\(t^2\left(t+1\right)\left(t-1\right)=56\)
\(\Leftrightarrow t^2\left(t^2-1\right)=56\)
\(\Leftrightarrow t^4-t^2-56=0\)
\(\Leftrightarrow t^4-8t^2+7t^2-56=0\)
\(\Leftrightarrow t^2\left(t^2-8\right)+7\left(t^2-8\right)=0\)
\(\Leftrightarrow\left(t^2-8\right)\left(t^2+7\right)=0\)
\(\Leftrightarrow t=8\)(vì t2+7>0)
Do đó 64x2+112x+41=0
Tới đây bạn denta hoặc vi-ét nó ra
\(\Leftrightarrow x=-\frac{\sqrt{2^3}+7}{8}\)hoặc\(x=\frac{\sqrt{2^3}-7}{8}\)
\(\Leftrightarrow2\left(8x+7\right)^2\left(4x+3\right)\left(x+1\right)=3,5\times2=7\)
\(\Leftrightarrow\left(8x+7\right)^2\left[2\left(4x+3\right)\right]\left[8\left(x+1\right)\right]=7\times8\)
Đặt t=8x+7.Pt trở thành:
$t^2\left(t+1\right)\left(t-1\right)=56$t2(t+1)(t−1)=56
$\Leftrightarrow t^2\left(t^2-1\right)=56$⇔t2(t2−1)=56
$\Leftrightarrow t^4-t^2-56=0$⇔t4−t2−56=0
$\Leftrightarrow t^4-8t^2+7t^2-56=0$⇔t4−8t2+7t2−56=0
$\Leftrightarrow t^2\left(t^2-8\right)+7\left(t^2-8\right)=0$⇔t2(t2−8)+7(t2−8)=0
$\Leftrightarrow\left(t^2-8\right)\left(t^2+7\right)=0$⇔(t2−8)(t2+7)=0
$\Leftrightarrow t=8$⇔t=8(vì t2+7>0)
Do đó 64x2+112x+41=0
Tới đây bạn denta hoặc vi-ét nó ra
$\Leftrightarrow x=-\frac{\sqrt{2^3}+7}{8}$⇔x=−√23+78 hoặc$x=\frac{\sqrt{2^3}-7}{8}$x=√23−78
