\(a,12-2:\left(x+3\right)=9\)

\(\Leftrightarrow2:\left(x+3\right)=3\)

\(\Leftrightarrow x+3=\frac{2}{3}\)

\(\Leftrightarrow x=\frac{-7}{3}\)

12-2:(x+3)=9

=> 12 - 2x - 6 = 9

=> -2x = 9+6-12 = 3

=> x= -1,5

12xy - 50 = 7xy

=> 12xy -7xy= 50

=> 5xy=50

=> xy = 10

=> x = 10/y

\(b,12y-50=7y\)

\(\Leftrightarrow12y-7y=50\)

\(\Leftrightarrow5y=50\)

\(\Leftrightarrow y=10\)

theo đề ta có 

(3x+2)(x+3) = (0,5x+2)(5x+7) 

=> 3x^2 + 11x + 6 = 2,5x+13,5x+14 

..................

từ đây bn tự giải tìm x

Để (x-1)(ý+2)=7 suy ra (x-1)=7 hoac (y+2)=7

TH1:

(x-1)=7

x     = 7 + 1

x     = 8 

TH2:

(y+2)=7

y     = 7-2

y     = 5 

Vậy : x=5 và x=8

Giải:

a) Ta có:

\(\left(x^4\right)^2=\frac{x^{12}}{x^5}\left(x\ne0\right)\Leftrightarrow x^8=x^7\)

\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\left(x^7\ne0\right)\Leftrightarrow x=1\)

Vậy \(x=1\)

b) Ta có:

\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)

\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow\) \(\left[\begin{array}{}x^8=0\\x^2-25=0\end{array}\right.\)

\(\Leftrightarrow\) \(\left[\begin{array}{}x=0\\x=5\\x=-5\end{array}\right.\) Vậy...

ta có:

x-y+y-z+x+z=8+10+12

2x=30

x=30:2=15

y=15-8=7

z=7-10=-3

khi đó: x+y+z=15+7+(-3)=19

chờ, e bít làm

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