Bài 1:
a) Bạn xem lại đề
b)
\(x^3-1=0\)
\(\Leftrightarrow (x-1)(x^2+x+1)=0\)
Vì \(x^2+x+1=x^2+2.\frac{1}{2}x+(\frac{1}{2})^2+\frac{3}{4}=(x+\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}>0\)
\(\Rightarrow x^2+x+1\neq 0\)
Do đó: \(x-1=0\Rightarrow x=1\) là nghiệm duy nhất
Bài 2:
a) \((x^2-5x)^2+10(x^2-5x)+24=0\)
\(\Leftrightarrow (x^2-5x)^2+2.5(x^2-5x)+5^2-1=0\)
\(\Leftrightarrow (x^2-5x+5)^2-1=0\)
\(\Leftrightarrow (x^2-5x+5-1)(x^2-5x+5+1)=0\)
\(\Leftrightarrow (x^2-5x+4)(x^2-5x+6)=0\)
\(\Leftrightarrow (x-1)(x-4)(x-2)(x-3)=0\)
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-4=0\\ x-2=0\\ x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=1\\ x=4\\ x=2\\ x=3\end{matrix}\right.\)
b)
\((x+2)(x+3)(x-5)(x-6)=180\)
\(\Leftrightarrow [(x+2)(x-5)][(x+3)(x-6)]=180\)
\(\Leftrightarrow (x^2-3x-10)(x^2-3x-18)=180\)
\(\Leftrightarrow a(a-8)=180\) (đặt \(x^2-3x-10=a\) )
\(\Leftrightarrow a^2-8a+16-196=0\)
\(\Leftrightarrow (a-4)^2-14^2=0\)
\(\Leftrightarrow (a-4-14)(a-4+14)=0\Leftrightarrow (a-18)(a+10)=0\)
\(\Rightarrow a=18\) hoặc $a=-10$
+) Nếu $a=18$ thì \(x^2-3x-10=18\)
\(\Leftrightarrow x^2-3x-28=0\)
\(\Leftrightarrow (x-7)(x+4)=0\Rightarrow \left[\begin{matrix} x=7\\ x=-4\end{matrix}\right.\)
+) Nếu $a=-10$ thì \(x^2-3x-10=-10\Leftrightarrow x^2-3x=0\Leftrightarrow x(x-3)=0\)
\(\Leftrightarrow \left[\begin{matrix} x=0\\ x=3\end{matrix}\right.\)
Vậy pt có 4 nghiệm \(x\in \left\{7;-4;0;3\right\}\)
