a) x(x+1)+3(x+1)=0

⇌ (x+1)(x+3)=0

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)

b)3x(12x-4)-2x(18x+3)=0

⇒36x²-12x-36x²+6x=0

⇒ -6x = 0

⇒ x=0

<=> 3^x(2x+3)-9(2x+3)=0

<=> (2x+3)(3^x-9)=0

<=> 2x+3=0 => x=-3/2

Và: 3^x-9=0 => 3^x=9=3² => x=2

Đs: x=-3/2 và x=2

\(3^{x+1}+2x.3^x-18x-27=0\)

\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3^x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=2\end{matrix}\right.\)

Vậy ...............

3^x.3+2x.3^x-(18x+27)=0

=> 3^x(3+2x)-9.(3+2x)=0

=> (3^x-9).(3+2x)=0

=> \(\left[{}\begin{matrix}3^x-9=0\\3+2x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3^x=9=3^2\\2x=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

3^x+1+2X.3^x-18x+27=0

<=> 3^x(3+2x)-9(3+2x)=0

<=> (3+2x)(3^x-9)=0

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

\(\Leftrightarrow\orbr{\begin{cases}x=-1,5\\x=2\end{cases}}\)

Cảm ơn

\(3^{x+1}+2x+3^x-18x-27=0\)

<=> \(3^x\left(3+2x\right)-9\left(2x+3\right)=0\)

<=> \(\left(2x+3\right)\left(3^x-9\right)=0\)

<=>\(\orbr{\begin{cases}x=-\frac{3}{2}\\x=2\end{cases}}\)

vậy.......

3^{x + 1} + 2x .3^x - 18x - 27 = 0

<=> 3^x ( 3 + 2x ) - 9 ( 2x + 3 ) = 0

<=> ( 3^x - 9 ) ( 2x + 3 ) = 0

<=> \(\orbr{\begin{cases}3^x-9=0\\2x+3=0\end{cases}}\)<=>\(\orbr{\begin{cases}3^x=9\\2x=-3\end{cases}}\)

<=>\(\orbr{\begin{cases}3^x=3^2\\x=-\frac{3}{2}\end{cases}}\)<=>\(\orbr{\begin{cases}x=2\\x=-\frac{3}{2}\end{cases}}\)

\(a,2x^2-18x+28=0\)

\(\Leftrightarrow2\left(x^2-9x+14\right)=0\)

\(\Leftrightarrow x^2-9x+14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

\(b,\dfrac{x-2}{x^2-9}+\dfrac{3x-1}{x+3}=\dfrac{2x+1}{x-3}+1\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(3x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(2x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-1=0\)

\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3x^2-10x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+7x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)\(\Rightarrow x-2+3x^2-10x+3-2x^2-7x-3-x^2+9=0\)

\(\Leftrightarrow-16x+7=0\)

\(\Leftrightarrow-16x=-7\)

\(\Leftrightarrow x=\dfrac{7}{16}\left(tm\right)\)

\(VậyS=\left\{\dfrac{7}{16}\right\}\)

a: =>x^2-9x+14=0

=>(x-2)(x-7)=0

=>x=2 hoặc x=7

b: =>x-2+(3x-1)(x-3)=(2x+1)(x+3)+x^2-9

=>x-2+3x^2-9x-x+3=2x^2+7x+3+x^2-9

=>3x^2-9x+1=3x^2+7x-6

=>-16x=-7

=>x=7/16

a: \(\Leftrightarrow3^x\cdot3+2x\cdot3^x-18x-27=0\)

\(\Leftrightarrow3^x\left(2x+3\right)-9\left(2x+3\right)=0\)

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

=>x=2 hoặc x=-3/2

b: \(\Leftrightarrow\left|2x+5\right|\cdot\dfrac{1}{2}-\dfrac{5}{4}\cdot2\cdot\left|2x+5\right|+\dfrac{7}{3}\cdot4\cdot\left|2x+5\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left|2x+5\right|=\dfrac{1}{44}\)

=>2x+5=1/44 hoặc 2x+1=-1/44

=>x=-219/88 hoặc x=-221/88