\(4x^2-\left(12x-12\right)x+9\left(x-1\right)^2=0\)

\(\Leftrightarrow4x^2-2.2x.3\left(x-1\right)+\left(3x-3\right)^2=0\)

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

Trả lời:

4x² - ( 12x - 12 ) x + 9 ( x - 1 )² = 0

<=> ( 2x )² - 2.2x.3( x - 1 ) + [ 3 ( x - 1 ) ]² = 0

<=> [ 2x - 3 ( x - 1 ) ]² = 0

<=> ( 2x - 3x + 3 )² = 0

<=> ( 3 - x )² = 0 

<=> 3 - x = 0

<=> x = 3

Vậy x = 3 là nghiệm của pt.

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

TH1 : \(x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

TH2 : \(x^2+1=-2x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Trả lời:

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}}\)

Vậy x = 1; x = - 1 là nghiệm của pt.

<=> \(4x^2-12x^2+12x+9\left(x^2-2x+1\right)=0\) 

,<=> \(4x^2-12x^2+12x+9x^2-18x+9=0\)

<=> \(x^2-6x+9=0\)

<=> \(x=3\)

\(4x^2-\left(x-5\right)^2=0\)

\(\Leftrightarrow\left(2x-x+5\right)\left(2x+x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-5\right)=0\Leftrightarrow x=-5;x=\frac{5}{3}\)

4x² - ( x - 5 )² = 0

<=> ( 2x )² - ( x - 5 )² = 0

<=> ( x + 5 )( 3x - 5 ) = 0

<=> x = -5 hoặc x = 5/3

25x⁴ - 10x³ + x² = 0

<=> x²(25x² - 10x + 1) = 0

<=> x²(5x - 1)² = 0

<=> \(\orbr{\begin{cases}x^2=0\\\left(5x-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{5}\end{cases}}\) 

Trả lời:

\(25x^4-10x^3+x^2=0\)

\(\Leftrightarrow x^2\left(25x^2-10x+1\right)=0\)

\(\Leftrightarrow x^2\left(5x-1\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{5}\end{cases}}}\)

Vậy x = 0; x = 1/5 là nghiệm của pt.

\(10x^4-15x^3=0\)

\(\Leftrightarrow5x^3\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)

Trả lời:

\(10x^4-15x^3=0\)

\(\Leftrightarrow5x^3\left(2x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x^3=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy x = 0; x = 3/2 là nghiệm của pt.

\(9x^2+6x^2+x=0\)

\(15x^2+x=0\)

\(x\left(15x+1\right)=0\)

\(\orbr{\begin{cases}x=0\left(TM\right)\\15x=-1\end{cases}\orbr{\begin{cases}x=0\left(TM\right)\\x=-\frac{1}{15}\left(TM\right)\end{cases}}}\)

Trả lời:

\(9x^2+6x^2+x=0\)

\(\Leftrightarrow x\left(9x+6x+1\right)=0\)

\(\Leftrightarrow x\left(15x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\15x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{15}\end{cases}}}\)

Vậy x = 0; x = - 1/15 là nghiệm của pt.

e can gap ak :((

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

\(\left(x^3-1\right)-3x\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)=0\)

\(\left(x-1\right)\left(x^2+x+1-3x\right)=0\)

\(\orbr{\begin{cases}x-1=0\\x^2-2x+1=0\end{cases}\orbr{\begin{cases}x=1\\\left(x-1\right)^2=0\end{cases}\orbr{\begin{cases}x=1\\x=1\end{cases}< =>x=1\left(TM\right)}}}\)

Trả lời:

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

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy x = 1 là nghiệm của pt.