3x+2 / 5x+7  = 3x-1 / 5x + 1

=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7) 

=> 3x ( 5x+1) + 2( 5x+1) = 3x( 5x+7) - 5x - 7

15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7

15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2

13x - 16x = -9

-3x = -9

x = 3

Vậy x=3

Đúng 1000%

đề bài là j

3x+2 / 5x+7  = 3x-1 / 5x + 1

=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7) 

=> 3x ( 5x+1) + 2( 5x+1) = 3x( 5x+7) - 5x - 7

15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7

15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2

13x - 16x = -9

-3x = -9

x = 3

\(2x^2-5x^3+2x^3y^3=x^2\left(2-5x+2y^3\right)\)

\(2x^2-5x^3+2x^3y^3=x^2\left(2-5x+2xy^3\right)\)

2x2−5x3+2x3y3=x2(2−5x+2y3) đó tui ko có số nhỏ nên đành write số to

\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}-\dfrac{5x+4}{3}>3\)

=>4/5x+3/5-6/7x+2/7-5/3x-4/3>3

=>-181/105x>362/105

=>x<-2

\(x^2+5x+6=0\)

\(\Leftrightarrow\left(x^2+2x\right)+\left(3x+6\right)=0\)

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

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

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

x^2+2x+3x+6=0

x(x+2)+3(x+2)=0

(x+3)(x+2)=0

=> x= - 3 hoac -2

\(x^2+5x+6=0\)

<=> \(x^2+2x+3x+6=0\)

<=>\(\left(x^2+2x\right)+\left(3x+6\right)=0\)

<=> \(x\left(x+2\right)+3\left(x+2\right)=0\)

<=> \(\left(x+2\right)\left(x+3\right)=0\)

<=> x+2=0 hoặc x+3=0

<=> x= -2 hoặc x=-3

Vậy x=-2 ,x=-3

\(x^2+5x+6=0\)

\(\Rightarrow x^2+2x+3x+6=0\)

\(\Rightarrow x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+3\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}-2\\-3\end{array}\right.\)

Vậy x = -2 và x = -3

Ta có: \(x^2+5x+6=0\)

<=> \(\left(x^2+2x\right)+\left(3x+6\right)=0\)

<=> \(\left(x+2\right)\left(x+3\right)=0\)

<=> \(\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)

Vậy x\(\in\left\{-3;-2\right\}\)

x² + 5x + 6 = 0

⇒x²+ 2x + 3x + 6 = 0

⇒x ( x + 2 ) + 3 ( x + 2 ) = 0

⇒( x + 2 ) ( x + 3 ) = 0

⇒\(\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\end{array}\right.\)

Vậy x = -2 và x = -3

\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)

A'=\(\dfrac{\dfrac{5x}{2\sqrt{x-9}}-5\sqrt{x-9}}{\left(5x^2\right)}\)

\(A'=0\rightarrow5x=10\left(x-9\right)\)

            \(\rightarrow x=18\)

\(MaxA=\dfrac{1}{30}\) khi \(x=18\)

\(A=\dfrac{2.3\sqrt{x-9}}{30x}\le\dfrac{3^2+x-9}{30x}=\dfrac{1}{30}\)

\(A_{max}=\dfrac{1}{30}\) khi \(\sqrt{x-9}=3\Leftrightarrow x=18\)