a) \(x^2+2x+9=\left(x^2+2x+1\right)+8=\left(x+1\right)^2+8\)
Ta có :
\(\left(x+1\right)^2\ge0\)
\(\Rightarrow\left(x+1\right)^2+8\ge8>0\)
Do đó đa thức vô nghiệm.
Vậy...
b) \(y^2-y+1=\left(y^2-2.\frac{1}{2}y+\frac{1}{4}\right)+\frac{3}{4}=\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(y-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
c) \(2y^2-2y+4\)
\(=2y^2-2y+\frac{1}{2}+\frac{7}{2}\)
\(=2\left(y^2-2.\frac{1}{2}.y+\frac{1}{4}\right)+\frac{7}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Ta có :
\(\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\ge\frac{7}{2}>0\)
Do đó đa thức vô nghiệm
Vậy...
d) \(3x^4+x^2+2\)
\(=2x^4+\left(x^4+2.\frac{1}{2}x^2+\frac{1}{4}\right)+3\)
\(=2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\)
Ta có :
\(\left(x^2\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2\ge0\)
\(\left(x^2+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow2\left(x^2\right)^2+\left(x^2+\frac{1}{2}\right)^2+3\ge3>0\)
Do đó đa thức vô nghiệm.
Vậy ...
e) \(x^2+x+1=\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có :
\(\left(x+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
Do đó đa thức vô nghiệm.
Vậy ...
f) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)\)
\(=\left(x-5\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=5\\x=1\end{cases}.}\)
g) \(x^3-x^2+2\)
\(=x^3-x^2+2x-2x+2\)
\(=\left(x^3-x\right)-\left(x^2-x\right)-2\left(x-1\right)\)
\(=x\left(x^2-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-x\left(x-1\right)-2\left(x-1\right)\)
\(=\left[x\left(x+1\right)-x-2\right]\left(x-1\right)\)
\(=\left(x^2+x-x-2\right)\left(x-1\right)\)
\(=\left(x^2-2\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-2=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x\in\left\{-\sqrt{2};\sqrt{2}\right\}\\x=1\end{cases}}.\)
