a) \(x\left(5+3x\right)-\left(x+1\right)\left(3x-2\right)=12\)

\(5x+3x^2-3x^2+2x-3x+2=12\)

\(4x=10\)

\(x=\frac{5}{2}\)

vậy \(x=\frac{5}{2}\)

\(13x\left(x-8\right)-x+8=0\)

\(13x\left(x-8\right)-\left(x-8\right)=0\)

\(\left(13x-1\right)\left(x-8\right)=0\)

\(\Rightarrow\orbr{\begin{cases}13x-1=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)

 vậy \(\orbr{\begin{cases}x=\frac{1}{13}\\x=8\end{cases}}\)

\(3^x+3^{x+1}=3^8.2+2.3^8.2017^0\)
\(3^x+3^x.3=3^8.2^2\)
\(3^x=3^8.2^2:3\)
\(3^x=3^7.2^2\)
 

3x+3x+1=38.2+2.38.201703x+3x+1=38.2+2.38.20170

\(3^x=8\)

\(\Rightarrow log_3\left(3^x\right)=log_38\)

\(\Rightarrow x=log_38=3log_32\)

a,x-(20-3x)=x+7

=>x-20+3x=x+7

=>x-20+3x-x=7

=>x+3x-x=7+20

=>3x=27

=>x=27:3

=>x=9

bai toan lop 6 nay kho qua ?

mik mới học lớp 5 thôi!

3x+8 = 3(x+2) +2 

=> 2 chia hết cho x+2 => x+2 =1;=> x =-1

                                      x+2 =2 => x =0

                                     x+2 =-1 => x =-3

                                    x+2 = -3 => x =-5

x( x + 3) + 3x + 8

\(x\left(x+3\right)+x+2x+8=0\)

\(x\left(x+4\right)+2\left(x+4\right)=0\)

\(\left(x+2\right).\left(x+4\right)=0\)

TH1:

\(x+2=0\)

\(x=0-2\)

\(x=-2\)

TH2: 

\(x+4=0\)

\(x=0-4\)

\(x=-4\)

Vậy \(x\in\left\{-2;-4\right\}\)

câu cuối là d)

a) 

\(|3x+1|=4\)

\(\Rightarrow\orbr{\begin{cases}3x+1=4\\3x+1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=4-1\\3x=-4-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\div3\\x=-5\div3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1,6667\end{cases}}\)

Vậy x = 1

a: \(x^3-4x^2-x+4=0\)

=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)

=>\(\left(x+3\right)\left(x^2-4\right)=0\)

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)