a: Đặt \(a=x^2+x\) (ĐK: a>=-1/4)
Phương trình sẽ trở thành: \(a^2+4a-12=0\)
=>\(a^2+6a-2a-12=0\)
=>a(a+6)-2(a+6)=0
=>(a+6)(a-2)=0
=>\(\left[\begin{array}{l}a+6=0\\ a-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=-6\left(loại\right)\\ a=2\left(nhận\right)\end{array}\right.\)
=>\(x^2+x=2\)
=>\(x^2+x-2=0\)
=>(x+2)(x-1)=0
=>\(\left[\begin{array}{l}x+2=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-2\\ x=1\end{array}\right.\)
b: \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\)
=>\(\left(x^2-4\right)\left(x^2-10\right)=72\)
Đặt \(a=x^2\left(a\ge0\right)\)
Phương trình sẽ trở thành:
(a-4)(a-10)=72
=>\(a^2-14a+40-72=0\)
=>\(a^2-14a-32=0\)
=>(a-16)(a+2)=0
=>\(\left[\begin{array}{l}a-16=0\\ a+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}a=16\left(nhận\right)\\ a=-2\left(loại\right)\end{array}\right.\)
=>\(x^2=16\)
=>\(\left[\begin{array}{l}x=4\\ x=-4\end{array}\right.\)
c: \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
=>\(\left(x-1\right)\left(x+5\right)\left(x-3\right)\left(x+7\right)=297\)
=>\(\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\) (1)
Đặt \(b=x^2+4x\) (ĐKXĐ: b>=-4)
(1) sẽ trở thành:
\(\left(b-5\right)\left(b-21\right)=297\)
=>\(b^2-26b+105-297=0\)
=>\(b^2-26b-192=0\)
=>(b-32)(b+6)=0
=>\(\left[\begin{array}{l}b-32=0\\ b+6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}b=32\left(nhận\right)\\ b=-6\left(loại\right)\end{array}\right.\)
=>b=32
=>\(x^2+4x=32\)
=>\(x^2+4x-32=0\)
=>(x+8)(x-4)=0
=>\(\left[\begin{array}{l}x+8=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-8\\ x=4\end{array}\right.\)
