Phần e)
\((x+2)(x+3)(x-7)(x-8)=144\)
\(\Leftrightarrow [(x+2)(x-7)][(x+3)(x-8)]=144\)
\(\Leftrightarrow (x^2-5x-14)(x^2-5x-24)=144\)
Đặt \(x^2-5x-24=a\). PT trở thành:
\((a+10)a=144\)
\(\Leftrightarrow a^2+10a=144\)
\(\Leftrightarrow (a+5)^2=169\)
\(\Leftrightarrow \left[\begin{matrix} a+5=13\rightarrow a=8\\ a+5=-13\rightarrow a=-18\end{matrix}\right.\)
Nếu \(a=8\Leftrightarrow x^2-5x-24=8\Leftrightarrow x^2-5x-32=0\)
\(\Leftrightarrow x=\frac{5\pm 3\sqrt{17}}{2}\)
Nếu \(a=-18\Rightarrow x^2-5x-24=-18\)
\(\Leftrightarrow x^2-5x-6=0\Leftrightarrow (x+1)(x-6)=0\Leftrightarrow x=-1\) hoặc \(x=6\)
Vậy..........
Phần f)
ĐKXĐ: \(x\geq \frac{1}{2}\)
\(2x+8\sqrt{2x-1}=21\)
\(\Leftrightarrow (2x-1)+8\sqrt{2x-1}+16=36\)
\(\Leftrightarrow (\sqrt{2x-1}+4)^2=36\)
\(\Leftrightarrow \left[\begin{matrix} \sqrt{2x-1}+4=6\\ \sqrt{2x-1}+4=-6\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} \sqrt{2x-1}=2\\ \sqrt{2x-1}=-10<0(\text{vô lý})\end{matrix}\right.\)
\(\Rightarrow \sqrt{2x-1}=2\Rightarrow 2x-1=4\Rightarrow x=\frac{5}{2}\) (thỏa mãn)
Vậy \(x=\frac{5}{2}\)
Phần i)
\(2x^2-3-4(x-1)=0\)
\(\Leftrightarrow 2x^2-4x+1=0\)
\(\Leftrightarrow 2(x^2-2x+1)-1=0\)
\(\Leftrightarrow 2(x-1)^2=1\Leftrightarrow \left[\begin{matrix} x-1=\frac{1}{\sqrt{2}}\\ x-1=-\frac{1}{\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow \left[\begin{matrix} x=1+\frac{\sqrt{2}}{2}\\ x=1-\frac{\sqrt{2}}{2}\end{matrix}\right.\)