\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow\left(144x^2+168x+49\right)\left(6x^2+7x+2\right)=3\)
\(\Leftrightarrow\left(144x^2+168x+49\right).24.\left(6x^2+7x+2\right)=3.24\)
\(\Leftrightarrow\left(144x^2+168x+49\right)\left(144x^2+168x+48\right)=72\) (*)
-Đặt \(144x^2+168x+49=a\)
(*) \(\Leftrightarrow a\left(a-1\right)=72\)
\(\Leftrightarrow a^2-a-72=0\)
\(\Leftrightarrow a^2-9a+8a-72=0\)
\(\Leftrightarrow a\left(a-9\right)+8\left(a-9\right)=0\)
\(\Leftrightarrow\left(a-9\right)\left(a+8\right)=0\)
\(\Leftrightarrow a-9=0\) hay \(a+8=0\)
\(\Leftrightarrow a=9\) hay \(a=-8\)
\(\Leftrightarrow144x^2+168x+40=0\) hay \(144x^2+168x+57=0\)
\(\Leftrightarrow144x^2+48x+120x+40=0\) hay \(144x^2+168x+49+8=0\)
\(\Leftrightarrow48x\left(\dfrac{1}{3}x+1\right)+40\left(\dfrac{1}{3}x+1\right)=0\) hay \(\left(12x+7\right)^2+8=0\) (vô nghiệm vì \(\left(12x+7\right)^2+8\ge8\))
\(\Leftrightarrow8\left(\dfrac{1}{3}x+1\right)\left(6x+5\right)=0\)
\(\Leftrightarrow\dfrac{1}{3}x+1=0\) hay \(6x+5=0\)
\(\Leftrightarrow x=\dfrac{-1}{3}\) hay \(x=\dfrac{-5}{6}\)
-Vậy \(S=\left\{\dfrac{-1}{3};\dfrac{-5}{6}\right\}\)
\(144x^2+168x+49=a\)
giúp mình bài này với ạ,ghi từng bước giúp mình lun ạ