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

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

a) \(x^3-16x=0\)

 ⇔\(x\left(x^2-16\right)=0\)

 ⇒\(x=0\) hoặc \(x^2-16=0\)

\(TH_1:x=0\)

\(TH_2:x^2-16=0\) ⇔ \(x^2=16\) ⇔ \(x=\pm4\)

             Vậy \(x\in\left\{0;\pm4\right\}\)

b) \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

⇒ \(2x+1=x-1\)

⇒ \(2x+2=x\)

⇒ \(2\left(x+1\right)=x\) ⇒ x = -2 

        Vậy x = -2

\(63x^2-16x+1=0\)

\(\Leftrightarrow63x^2-9x-7x+1=0\)

\(\Leftrightarrow9x\left(7x-1\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow\left(9x-1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}9x-1=0\\7x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=\frac{1}{7}\end{cases}}}\)

Bài làm:

Ta có: \(63x^2-16x+1=0\)

\(\Leftrightarrow\left(63x^2-9x\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow9x\left(7x-1\right)-\left(7x-1\right)=0\)

\(\Leftrightarrow\left(7x-1\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x-1=0\\9x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=\frac{1}{7}\end{cases}}\)

63x² - 16x + 1 = 0

<=> 63x² - 9x - 7x + 1 = 0

<=> 9x( 7x - 1 ) - 1( 7x - 1 ) = 0

<=> ( 7x - 1 )( 9x - 1 ) = 0

<=> \(\orbr{\begin{cases}7x-1=0\\9x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{7}\\x=\frac{1}{9}\end{cases}}\)

trả lời giúp mình 

\(16x^3-16x^4+4x-8x^2-1=0\)

<=>  \(-16x^4-4x^2+16x^3+4x-4x^2-1=0\)

<=>  \(-4x^2\left(4x+1\right)+4x\left(4x^2+1\right)-\left(4x^2+1\right)=0\)

<=>  \(-\left(4x^2+1\right)\left(4x^2-4x+1\right)=0\)

<=>  \(-\left(4x^2+1\right)\left(2x-1\right)^2=0\)

<=>   \(2x-1=0\) (do  4x² + 1 > 0 )

<=>  \(x=\frac{1}{2}\)

cảm ơn bạn nha 

Ta có : 63x² + 16x + 1 = 0

=> 63x² + 9x + 7x + 1 = 0

=> 9x(7x + 1) + (7x + 1) = 0

=> (9x + 1)(7x + 1) = 0

=> \(\orbr{\begin{cases}9x+1=0\\7x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{9}\\x=-\frac{1}{7}\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{9};-\frac{1}{7}\right\}\)

\(63x^2+16x+1=0\)

\(\Leftrightarrow63x^2+7x+9x+1=0\)

\(\Leftrightarrow7x\left(9x+1\right)+\left(9x+1\right)=0\)

\(\Leftrightarrow\left(7x+1\right)\left(9x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}7x+1=0\\9x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{7}\\x=-\frac{1}{9}\end{cases}}\)

63x² + 16x + 1 = 0

<=> 63x² + 9x + 7x + 1 = 0

<=> 9x( 7x + 1 ) + 1( 7x + 1 ) = 0

<=> ( 7x + 1 )( 9x + 1 ) = 0

<=> \(\orbr{\begin{cases}7x+1=0\\9x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{7}\\x=-\frac{1}{9}\end{cases}}\)

a: Ta có: \(x\left(2-x\right)+x^2+x=7\)

\(\Leftrightarrow2x-x^2+x^2+x=7\)

\(\Leftrightarrow3x=7\)

hay \(x=\dfrac{7}{3}\)

b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

a) \(x^3-16x=0\)

\(\Leftrightarrow x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{16}=\pm4\end{cases}}\)

Vậy \(x\in\left\{0;\pm4\right\}\)

b) \(x^2-6x+9=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

\(a)x^3-16x=0=>x\left(x^2-16\right)=0\)

\(=>x\left(x-4\right)\left(x+4\right)=0\)

Suy ra x=0 hoặc x-4=0 hoặc x+4=0

=> x=0 hoặc x=4 hoặc x=-4

b)\(x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)

d) <=>x²-5x-x+5=0

<=>x(x-5)-(x-5)=0

<=>(x-5)(x-1)=0

<=>x=5 hoặc x=1

thank nha