Question

Chứng minh các phương trình sau vô nghiệm :

a, \(x^4+2x^2-6x+7=0\)

b,\(|x-2|+|x^2-4x+3|=0\)

Answer 1

a ) \(x^4+2x^2-6x+7=0\)

\(\Leftrightarrow x^4-2x^2+1+4x^2-6x+6=0\)

\(\Leftrightarrow\left(x^2-1\right)^2+4x^2-2.2x.\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x^2-1\right)^2+\left(2x-\dfrac{3}{2}\right)^2=-\dfrac{15}{4}\left(VL\right)\)

=> PTVN

b ) \(\left|x-2\right|\ge0;\left|x^2-4x+3\right|\ge0\forall x\)

\(\Rightarrow\left|x-2\right|+\left|x^2-4x+3\right|\ge0\forall x\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\x^2-4x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\end{matrix}\right.\)

Lại có : \(\left|x-2\right|+\left|x^2-4x+3\right|=0\) ( * )

Thay \(x=2\) vào ( * ) , ta có :

\(0+\left|2^2-4.2+3\right|=0\)

\(\Leftrightarrow0+\left|4-8+3\right|=0\Leftrightarrow0+1=0\Leftrightarrow1=0\)

( ***** ) (1)

Tương tư thay \(x=1\) \(\Rightarrow1=0\left(VL\right)\) (2)

thay \(x=3\Rightarrow1=0\left(L\right)\) (3)

Từ (1) ; (2) ; (3) \(\Rightarrow PTVN\)