a/ \(x^3+2x^2-8x+5=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(3x^2-3x\right)-\left(5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+3x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-\sqrt{29}\right)\left(2x+3+\sqrt{29}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+3-\sqrt{29}=0\\2x+3+\sqrt{29}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-3+\sqrt{29}}{2}\\x=\dfrac{-3-\sqrt{29}}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm
\(x=\left\{1;\dfrac{-3+\sqrt{29}}{2};\dfrac{-3-\sqrt{29}}{2}\right\}\)
b/ \(x^3-2x^2+1=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(x^2-x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1-\sqrt{5}\right)\left(2x-1+\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-1-\sqrt{5}=0\\2x-1+\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
Vậy phương trình đã cho có nghiệm \(x=\left\{1;\dfrac{1+\sqrt{5}}{2};\dfrac{1-\sqrt{5}}{2}\right\}\)
c/ \(x^2-x+1=0\) (1)
Ta có: \(x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}\) = \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta luôn có: \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) với mọi x
\(\Leftrightarrow x^2-x+1>0\) với mọi x (2)
Ta thấy (1) và (2) mâu thuẫn \(\Rightarrow\) Phương trình đã cho vô nghiệm
Câu 1
a, \(x^3_{ }+2x^2-8x+5=x^3-x^2+3x^2-3x-5x+5=x^2\left(x-1\right)+3x\left(x-1\right)-5\left(x-1\right)=\left(x^2+3x-5\right)\left(x-1\right)\)\(x^3+2x^2-8x+5=0.\left\{{}\begin{matrix}x^2+3x-5=0\\x-1=0\end{matrix}\right.< =>\left\{{}\begin{matrix}\left(x+1,5\right)^2=7,25\\x=1\end{matrix}\right.< =>\left\{{}\begin{matrix}x_1-1,5+\sqrt{7,25}_{ }_{ }_{ }x_2=-1,5-\sqrt{7,25}\\x=1\end{matrix}\right._{ }\)
Nhớ tick tiếp mình nha
b, \(x^3-2x^2+1=x^3-x^2-\left(x^2-1\right)=x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+x+1\right)\)
