hình như x = 50 hay 40

ko chắc bn ạ đừng chép bài mình

rõ ràng hơn đi , mk cần cả bài giải

b,|1-2x|+|5-2x|=4

==>|1-2x|+|5-2x|=|1-2x|+|2x-5|

==>|1-2x|+|2x-5| \(\geq\) |1-2x+2x-5|=|-4|=4

A \(\geq\) 4 khi \(\hept{\begin{cases}1-2x\\2x-5\end{cases}}\)\(\geq\) 0 ==>1/2\(\leq \) x \(\leq \) 5/2

\(ĐKXĐ:x\ne2\)

\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)  

\(\Leftrightarrow\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{x^3-8}\)

\(\Rightarrow9x^2+6x-12=3x^2+6x+12\)

\(\Leftrightarrow9x^2-3x^2+6x-6x-12-12=0\)

\(\Leftrightarrow6x^2-24=0\)

\(\Leftrightarrow6\left(x^2-4\right)=0\)

\(\Leftrightarrow x^2-4=0\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=\pm2\) (loại x = 2)

vậy x = -2

\(\frac{9x^2}{x^3-8}+\frac{6}{x^2+2x+4}=\frac{3}{x-2}\)

=>\(\frac{9x^2}{x^3-8}+\frac{6\left(x-2\right)}{x^3-8}-\frac{3\left(x^2+2x+4\right)}{x^3+8}=0\)

=>\(9x^2+6x-12-3x^2-6x-24=0\)

=>\(6x^2-36\)\(6x^2-6\)

=>\(\left(6x-6\right)\left(6x+6\right)\)

=> \(6\left(x-1\right)6\left(x+1\right)\)

=>\(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

#kenz

\(ĐKXĐ:x\ne2\)

\(pt\Leftrightarrow\frac{9x^2+6x-12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{x^3-8}\)

\(\Leftrightarrow9x^2+6x-12=3x^2+6x+12\)

\(\Leftrightarrow6x^2=24\Leftrightarrow x=\pm2\)

Vậy x = -2

a) \(\left(-4\right)^{\left(x+3\right)}=\left(-8\right)^{-2}\Rightarrow\left(-2\right)^{\left(2x+4\right)}=\left(-2\right)^{-6}\Rightarrow2x+4=-6\Rightarrow x=-5\)

Vì (2x-1)^6=(2x-1)^8

(2x-1)^8-(2x-1)^6=0

(2x-1)^6[(2x-1)^2-1)]=0

th1 (2x-1)^6 suy ra 2x-1=0 suy ra x=1/2

th2 (2x-1)^2-1=0

(2x-1)^2=1

suy ra 2x-1 bằng 1;-1

th1 2x-1=1 suy ra x=1

2x-1=-1 suy ra x=0

hay đấy nhưng tớ ko giải đâu

chuẩn cmnr