Question

(3x - 1) ( x + 3) + 9x² - 1= 0

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

Giúp mình với mình cần gấp

 

Answer 1

( 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

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

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

\(\Leftrightarrow4x< 11\)

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

Answer 3

(3x-1)(x+3)+9x²-1=0

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

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

<=> (3x-1)(-2x+2)=0

\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\-2x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

b) \(x+\frac{1}{3}>3x-\frac{2}{5}\)

\(\Leftrightarrow x+\frac{1}{3}-3x+\frac{2}{5}>0\)

\(\Leftrightarrow-2x+\frac{11}{15}>0\)

\(\Leftrightarrow2x< \frac{11}{15}\Leftrightarrow x< \frac{22}{15}\)

Answer 4

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

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

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

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

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

b, \(\frac{x+1}{3}>3x-\frac{1}{5}\)

\(\Leftrightarrow\frac{5x+5}{15}>\frac{45x}{15}-\frac{3}{15}\Leftrightarrow5x+5>45x-3\)

Đến đây dễ, làm nốt nhé !