Ta có:\(8x^2+10x+3=\left(8x^2+6x\right)+\left(4x+3\right)\)
\(=2x\left(4x+3\right)+\left(4x+3\right)\)
\(=\left(2x+1\right)\left(4x+3\right)\)
\(4x^2+7x+3=\left(4x^2+4x\right)+\left(3x+3\right)\)
\(=4x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(4x+3\right)\)
\(ĐKXĐ:x\ne-1,x\ne\frac{-3}{4}\)
\(8x^2+10x+3=\frac{1}{4x^2+7x+3}\)
<=>\(\left(8x^2+10x+3\right)\left(4x^2+7x+3\right)=1\)
<=>\(\left(2x+1\right)\left(4x+3\right)\left(x+1\right)\left(4x+3\right)=1\)
<=>\(\left(2x+1\right)\left(4x+3\right)^2\left(x+1\right)=1\)
<=>\(\left(4x+2\right)\left(4x+3\right)^2\left(4x+4\right)=8\)
(Nhân cả 2 vế với 8)
<=>\(\left[\left(4x+2\right)\left(4x+4\right)\right]\left(4x+3\right)^2=8\)
<=>\(\left(16x^2+24x+8\right)\left(16x^2+24x+9\right)=8\)
Đặt \(16x^2+24x+8.5=y\)
\(ĐK:y>-0.5\)
(Vì \(16x^2+24x+8.5=\left(4x+3\right)^2-0.5>-0.5\)với mọi x thỏa mãn ĐKXĐ)
Phương trình trở thành:
(y-0.5)(y+0.5)=8
<=>\(y^2-0.25=8\)
<=>\(y^2=8.25\)
<=>\(\orbr{\begin{cases}y=\frac{\sqrt{33}}{2}\left(\text{thỏa mãn}\right)\\y=\frac{-\sqrt{33}}{2}\left(\text{loại}\right)\end{cases}}\)
Với \(y=\frac{\sqrt{33}}{2}\)
Ta có:\(16x^2+24x+8.5=\frac{\sqrt{33}}{2}\)
<=>\(32x^2+48x+17-\sqrt{33}=0\)
<=>\(\left(x\sqrt{33}+3\sqrt{2}\right)^2=\sqrt{33}+1\)
<=>\(\orbr{\begin{cases}x\sqrt{33}+3\sqrt{2}=\sqrt{\sqrt{33}+1}\\x\sqrt{33}+3\sqrt{2}=-\sqrt{\sqrt{33+1}}\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{\sqrt{\sqrt{33}+1}-3\sqrt{2}}{\sqrt{33}}\\x=\frac{-\sqrt{\sqrt{33}+1}-3\sqrt{2}}{\sqrt{33}}\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{\sqrt{33\sqrt{33}+33}-3\sqrt{66}}{33}\left(\text{thỏa mãn ĐKXĐ}\right)\\x=\frac{-\sqrt{33\sqrt{33}+33}-3\sqrt{66}}{33}\left(\text{thỏa mãn ĐKXĐ}\right)\end{cases}}\)
(Kết luận: Vậy tập nghiệm của phương trình đã cho là...)
