64x^3 + 1 - 64x^3 + 80x =17

80x                              =16

   x                               =3/10

64x^3 + 1 -64x^3 + 80x = 17

80x                             = 16

   x                              = 3/10

64x^3 + 1 - 64x^3 + 80x = 17

80x                               = 16

   x                                = 3/10

a) \(4x^2+16x+3=0\)

\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)

b) \(7x^2+16x+2=1+3x^2\)

\(4x^2+16x+1=0\)

\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)

c) \(4x^2+20x+4=0\)

\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)

\(\Leftrightarrow x^2+5x+1=0\)

\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)

Phương trình có 2 nghiệm

\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)

(x - 1)² - (x + 2)(x - 5) + (4x - 1)² = (-x - 1)(2 - 16x)

=> x² - 2x + 1 - x² + 5x - 2x + 10 + 16x² - 8x + 1 = -2x + 16x² -2 + 16x

=> 16x² - 7x + 12 = 14x - 2  + 16x²

=> -21x = -14

=> 21x = 14

=> x = 2/3

\(\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)\)

\(\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)

\(\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0\)

\(\Rightarrow-21x+14=0\)

\(\Rightarrow-21x=-14\)

\(\Rightarrow x=\frac{2}{3}\)

\(x^2-2x+1-\left(x^2-3x-10\right)+16x^2-8x+1=16x^2+14x-2\)    

\(x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)         

\(-21x-14=0\) 

\(-21x=14\)       

\(x=\frac{-2}{3}\)