1, f(x)=x3-x2+x-1
\(f\left(x\right)=x^2\left(x-1\right)+\left(x-1\right)=\left(x^2+1\right)\left(x-1\right)\)
f(x) = 0 \(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+1=0\\x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=-1\\x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\in\phi\\x=1\end{array}\right.\)
\(\Leftrightarrow x=1\)
4, P(x)=x2+5x
\(\Rightarrow P\left(x\right)=x\left(x+5\right)\)
\(P\left(x\right)=0\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=0\\x+5=0\Rightarrow x=-5\end{array}\right.\)
6, Q(x)=3x2-4x
\(\Rightarrow Q\left(x\right)=x\left(3x-4\right)\)
\(Q\left(x\right)=0\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=0\\3x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x=4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{4}{3}\end{array}\right.\)7, H(x)= 5x5+10x
\(\Rightarrow H\left(x\right)=5x\left(x^4+2\right)\)\(H\left(x\right)=0\Leftrightarrow\)\(\left[\begin{array}{nghiempt}5x=0\\x^4+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^4=-2\end{array}\right.\)\(\Leftrightarrow x=0\)
Mình chỉ tìm nguyên nghiệm đc thôi k làm đc cách giải đâu
Ai lm đc thì lm đj, đừng ở đó lên mặt
