Question

[chi mai Nguyen]

(3x - 1) ( x + 3) + 9x² - 1= 0

x + 1/3 > 3x - 2/5

Giúp mình với mình cần gấp

Answer 1

[l҉o҉n҉g҉ d҉z҉]

( 3x - 1 )( x + 3 ) + 9x² - 1 = 0

<=> 3x² + 9x - x - 3 + 9x² - 1 = 0

<=> 12x² + 8x - 4 = 0

<=> 4( 3x² + 2x - 1 ) = 0

<=> 3x² + 2x - 1 = 0 

<=> 3x² + 3x - x - 1 = 0

<=> ( 3x² + 3x ) - ( x + 1 ) = 0

<=> 3x( x + 1 ) - 1( x + 1 ) = 0

<=> ( 3x - 1 )( x + 1 ) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Vậy S = { 1/3 ; -1 }

\(\frac{x+1}{3}>\frac{3x-2}{5}\)

\(\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Leftrightarrow5x+5>9x-6\)

\(\Leftrightarrow5x-9x>-6-5\)

\(\Leftrightarrow-4x>-11\)

\(\Leftrightarrow x< \frac{11}{4}\)

Answer 2

[Nguyễn Minh Đăng]

Bài làm:

a) \(\left(3x-1\right)\left(x+3\right)+9x^2-1=0\)

\(\Leftrightarrow3x^2+8x-3+9x^2-1=0\)

\(\Leftrightarrow12x^2+8x-4=0\)

\(\Leftrightarrow3x^2+2x-1=0\)

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

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

Vậy tập nghiệm của PT \(S=\left\{-1;\frac{1}{3}\right\}\)

b) \(\frac{x+1}{3}>\frac{3x-2}{5}\Leftrightarrow\frac{5\left(x+1\right)}{15}>\frac{3\left(3x-2\right)}{15}\)

\(\Rightarrow5x+5>9x-6\)

\(\Leftrightarrow4x< 11\)

\(\Rightarrow x< \frac{11}{4}\)