Question

Tìm x:

63x^2 - 16x + 1 = 0

Answer 1

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

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

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

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

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

Answer 2

Bài làm:

Ta có: \(63x^2-16x+1=0\)

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

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

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

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

Answer 3

63x² - 16x + 1 = 0

<=> 63x² - 9x - 7x + 1 = 0

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

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

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

Answer 4

                                  Bài làm :

Ta có :

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

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

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

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

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

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

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

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 5

Tìm x:

63x² - 16x + 1 = 0
<=> 63x² - 9x - 7x + 1 = 0
<=> 9x( 7x - 1 ) - 1( 7x - 1 ) = 0
<=> ( 7x - 1 )( 9x - 1 ) = 0

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