Tìm x biết: 5x^2 – x – 3 = 2x(x – 1) – 1 + x^2

2 bình luận về “Tìm x biết: 5x^2 – x – 3 = 2x(x – 1) – 1 + x^2”

1. Giải đáp:

    5x^{2}-x-3=2x.(x-1)-1+x^{2}

   => 5x^{2}-x-3=2x^{2}-2x-1+x^{2}

   => 5x^{2}-x-3=3x^{2}-2x-1

   => 5x^{2}-x-3-3x^{2}+2x+1=0

   => 2x^{2}+x-2=0

   => 2.[x^{2}+2.x.\frac{1}{2}+(\frac{1}{2})^{2}-\frac{7}{4}]=0

   =>(x+\frac{1}{2})^{2}=\frac{17}{4}

   => (x+\frac{1}{2})^{2}=(\pm\sqrt{\frac{17}{4}})^{2}

   => \(\left[ \begin{array}{l}x+\dfrac{1}{2}=\sqrt{\dfrac{17}{4}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{17}{4}}\end{array} \right.\) => \(\left[ \begin{array}{l}x=\dfrac{\sqrt{17}}{2}-\dfrac{1}{2}\\x=\dfrac{-\sqrt{17}}{2}-\dfrac{1}{2}\end{array} \right.\) => \(\left[ \begin{array}{l}x=\dfrac{\sqrt{17}-1}{2}\\x=\dfrac{-\sqrt{17}-1}{2}\end{array} \right.\) 

   => x=\frac{\pm\sqrt{17}-1}{2}

   Vậy x=\frac{\pm\sqrt{17}-1}{2}

   #DYNA

   Trả lời
2. 5x^2-x-3=2x(x-1)-1+x^2

   =>5x^2-x-3=2x^2-2x-1+x^2

   =>5x^2-x-3=3x^2-2x-1

   =>5x^2-x-3-3x^2+2x+1=0

   =>2x^2+x-2=0

   =>2(x^2+1/2 x-1)=0

   =>x^2+1/2 x-1=0

   =>x^2+2.x .1/4+(1/4)^2-17/16=0

   =>(x+1/4)^2=17/16

   TH1:(x+1/4)^2=( – $\dfrac{ \sqrt[]{17}}{4}$ )^2

   =>x+1/4=- $\dfrac{\sqrt[]{17}}{4}$ =>x= $\dfrac{-1-\sqrt[]{17}}{4}$ 

   TH2:(x+1/4)^2=( $\dfrac{ \sqrt[]{17}}{4}$ )^2

   =>x+1/4= $\dfrac{\sqrt[]{17}}{4}$ =>x= $\dfrac{-1+\sqrt[]{17}}{4}$ 

   =>x= $\dfrac{-1±\sqrt[]{17}}{4}$

    

   Trả lời