Giải các phương trình sau:
[TEX]a.(6x+7)^2(3x+4)(x+1)=1[/TEX]
[TEX]b.3(x^2+2x-1)^2-2(x^2+3x-1)^2+5x^2=0[/TEX]
[tex]a)(6x+7)^2(3x+4)(x+1)=1\\\Leftrightarrow (36x^2+84+49)(3x^2+7x+4)=1(*)[/tex]
Đặt [tex]3x^2+7x+4=a[/tex]
Khi đó [tex](*)\Leftrightarrow (12a+1)a=1[/tex]
...
[tex]b)3(x^2+2x-1)^2-2(x^2+3x-1)^2+5x^2=0\\\Leftrightarrow 2[(x^2+2x-1)^2-(x^2+3x-1)^2]+(x^2+2x-1)^2+5x^2=0\\\Leftrightarrow 2(-x)(2x^2+5x-2)+(x^2+2x-1)^2+5x^2=0\\\Leftrightarrow -2x^2-2x(2x^2+4x-2)+(x^2+2x-1)^2+5x^2=0\\\Leftrightarrow (x^2+2x-1)^2-4x(x^2+2x-1)+4x^2-x^2=0\\\Leftrightarrow [(x^2+2x-1)-2x]^2-4x^2=0\\\Leftrightarrow (x^2-1)^2-x^2=0\\\Leftrightarrow (x^2-x-1)(x^2+x-1)=0\\\Leftrightarrow ...[/tex]