sorry mk nhầm,từ dòng 4 fải lm như sau:
\(2x^2-4x=-10\)
\(\Leftrightarrow2x^2-4x+10=0\)
\(\Leftrightarrow2\left(x^2-2x+5\right)=0\)
\(\Leftrightarrow x^2-2x+5=0\)
\(\Leftrightarrow x^2-2x+1+4=0\)
\(\Leftrightarrow\left(x-1\right)^2+4=0\)
Vì: \(\left(x-1\right)^2+4\ge4>0\forall x\)
=>PT vô nghiệm
\(\left(2x-3\right)^2+\left(x+1\right)^2=3x\left(x-2\right)\)
\(\Rightarrow\left(2x-3\right)^2+\left(x+1\right)^2-3x\left(x-2\right)=0\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+3^2+x^2+2x+1-3x^2+6x=0\)
\(\Rightarrow4x^2-12x+9+x^2+2x+1-3x^2+6x=0\)
\(\Rightarrow2x^2-4x+10=0\)
\(\Rightarrow2\left(x^2-2x+5\right)=0\)
\(\Rightarrow x^2-2x+5=0\)
\(\Rightarrow x^2-2x+1+4=0\)
\(\Rightarrow\left(x-1\right)^2+4=0\)
Vì \(\left(x-1\right)^2\ge0\) với mọi x
\(4>0\)
\(\Rightarrow\left(x-1\right)^2+4>0\)
\(\Rightarrow\) Phương trình vô nghiệm
\(\left(2x-3\right)^2+\left(x+1\right)^2=3x\left(x-2\right)\)
\(\Leftrightarrow4x^2-12x+9+x^2+2x+1=3x^2-6x\)
\(\Leftrightarrow4x^2-12x+x^2+2x-3x^2+6x=-9-1\)
\(\Leftrightarrow2x^2-4x=-10\)
\(\Leftrightarrow2x^2-2x+10=0\)
\(\Leftrightarrow2\left(x^2-x+5\right)=0\)
\(\Leftrightarrow x^2-x+5=0\)
\(\Leftrightarrow x^2-x+\dfrac{1}{4}+\dfrac{19}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\)
Vì: \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\forall x\)
=> PT vô nghiệm
=.= hok tốt !!
