5x.5x.5x giúp mk vs
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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%
\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}-\dfrac{5x+4}{3}>3\)
=>4/5x+3/5-6/7x+2/7-5/3x-4/3>3
=>-181/105x>362/105
=>x<-2
\(x^2+5x+6=0\)
\(\Leftrightarrow\left(x^2+2x\right)+\left(3x+6\right)=0\)
\(\Leftrightarrow x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
\(x^2+5x+6=0\)
\(\Rightarrow x^2+2x+3x+6=0\)
\(\Rightarrow x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}-2\\-3\end{array}\right.\)
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\(\in\left\{-3;-2\right\}\)
\(A=\dfrac{\sqrt{x-9}}{5x}\left(ĐKx\ge9\right)\)
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 \(\sqrt{x-9}=3\Leftrightarrow x=18\)
Ta có : 5x.5x.5x=5.5.5.x.x.x=5^3.x^3=125.x^3
\(=125x^3\)