1: \(\left(2x^2+3x-1\right)^2-10x^2-15x+9=0\)
=>\(\left(2x^2+3x-1\right)^2-10x^2-15x+5+4=0\)
=>\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x-1\right)+4=0\)
=>\(\left(2x^2+3x-2\right)\left(2x^2+3x-5\right)=0\)
=>\(\left(2x^2+4x-x-2\right)\left(2x^2+5x-2x-5\right)=0\)
=>(x+2)(2x-1)(2x+5)(x-1)=0
=>\(\left[{}\begin{matrix}x+2=0\\2x-1=0\\2x+5=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)
2: \(\left(x^2-6x\right)^2-2\left(x-3\right)^2=81\)
=>\(\left[\left(x^2-6x+9\right)-9\right]^2-2\left(x-3\right)^2=81\)
=>\(\left[\left(x-3\right)^2-9\right]^2-2\left(x-3\right)^2=81\)(3)
Đặt \(c=\left(x-3\right)^2\left(c>=0\right)\)
(3) sẽ trở thành: \(\left(c-9\right)^2-2c=81\)
=>\(c^2-18c+81-2c-81=0\)
=>\(c^2-20c=0\)
=>c(c-20)=0
=>\(\left[{}\begin{matrix}c=0\\c=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x-3\right)^2=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-3=2\sqrt{5}\\x-3=-2\sqrt{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=2\sqrt{5}+3\\x=-2\sqrt{5}+3\end{matrix}\right.\)
3: \(\left(x^2-2x\right)^2-2\left(x-1\right)^2+2=0\)
=>\(\left(x^2-2x+1-1\right)^2-2\left(x-1\right)^2+2=0\)
=>\(\left[\left(x-1\right)^2-1\right]^2-2\left(x-1\right)^2+2=0\)(2)
Đặt \(b=\left(x-1\right)^2\left(b>=0\right)\)
(2) sẽ trở thành: \(\left(b-1\right)^2-2b+2=0\)
=>\(\left(b-1\right)^2-2\left(b-1\right)=0\)
=>(b-1)(b-3)=0
=>\(\left[{}\begin{matrix}b-1=0\\b-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=1\\b=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=1\\\left(x-1\right)^2=3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\)
4: \(\left(x^2-6x\right)^2-2\left(x-3\right)^2+2=0\)
=>\(\left(x^2-6x+9-9\right)^2-2\left(x-3\right)^2+2=0\)
=>\(\left[\left(x-3\right)^2-9\right]^2-2\left(x-3\right)^2+2=0\)(1)
Đặt \(a=\left(x-3\right)^2\)(ĐK: a>=0)
(1) sẽ trở thành: \(\left(a-9\right)^2-2a+2=0\)
=>\(a^2-18a+81-2a+2=0\)
=>\(a^2-20a+83=0\)
=>\(\left[{}\begin{matrix}a=10+\sqrt{17}\left(nhận\right)\\a=10-\sqrt{17}\left(nhận\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left(x-3\right)^2=10+\sqrt{17}\\\left(x-3\right)^2=10-\sqrt{17}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x-3=\sqrt{10+\sqrt{17}}\\x-3=-\sqrt{10+\sqrt{17}}\\x-3=\sqrt{10-\sqrt{17}}\\x-3=-\sqrt{10-\sqrt{17}}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\sqrt{10+\sqrt{17}}+3\\x=-\sqrt{10+\sqrt{17}}+3\\x=\sqrt{10-\sqrt{17}}+3\\x=-\sqrt{10-\sqrt{17}}+3\end{matrix}\right.\)
