Question

Tìm x: 

63x^2+16x+1=0

Answer 1

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

Answer 2

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

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

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

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

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

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

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

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

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

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

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

Answer 5

Bài làm 

\(63x^2+16x+1=0\Leftrightarrow\left(7x+1\right)\left(9x+1\right)=0\)

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