\(a,\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+2=0\)
\(\Leftrightarrow2\left(x+1\right)^2=-2\)
\(\Leftrightarrow\left(x+1\right)^2=-1\Rightarrow\) pt vô nghiệm
\(b,\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)-14x^2=0\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\)
\(\Leftrightarrow-8x=17\)
\(\Leftrightarrow x=\dfrac{-17}{8}\)
\(c,\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2+4x+4\right)-\dfrac{5}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+2\right)^2=\dfrac{5}{2}\)
\(\Rightarrow\left(x+2\right)^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x+2=-\sqrt{5}\\x+2=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)
a) \(\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\Leftrightarrow\left(\sqrt{2}x\right)^2+2.\sqrt{2}.\sqrt{2}x+\left(\sqrt{2}\right)^2+2=0\) \(\Leftrightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2=0\)
ta có : \(\left(\sqrt{2}x+\sqrt{2}\right)^2\ge0\Rightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2\ge2>0\forall x\)
\(\Rightarrow\) phương trình vô nghiệm
vậy phương trình vô nghiệm
b) \(\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\Leftrightarrow-8x=17\Leftrightarrow x=\dfrac{-17}{8}\)
vậy \(x=\dfrac{-17}{8}\)
c) \(\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}x^2-x+\dfrac{1}{2}\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x\right)^2+2.\sqrt{2}.\dfrac{\sqrt{2}}{2}x+\left(\sqrt{2}\right)^2-\dfrac{5}{2}=0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2-\dfrac{5}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2=\dfrac{5}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x+\sqrt{2}=\sqrt{\dfrac{5}{2}}\\\dfrac{\sqrt{2}}{2}x+\sqrt{2}=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x=\sqrt{\dfrac{5}{2}}-\sqrt{2}=\dfrac{\sqrt{10}-2\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2}x=-\sqrt{\dfrac{5}{2}}-\sqrt{2}=-\dfrac{\sqrt{10}+2\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{matrix}\right.\)
vậy \(x=-2+\sqrt{5};x=-2-\sqrt{5}\)
