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a) 2x + 6 = 12
2x = 6
x = 3
b) ( 3x + 3 ) - 5 = 13
3x + 3 = 18
3x = 15
x = 5
Vậy,...........
\(\left(2x\right)+6=12\) \(\left(3x+3\right)-5=13\)
\(2x=12-6\) \(3x+3=13+5\)
\(2x=6\) \(3x+3=18\)
\(x=6:2\) \(3x=18-3\)
\(x=3\) \(3x=15\Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2+6=8
12+6=20
25-23= 2
12-6= 6
k tớ k lại cho
^ _ ^ hi ủng hộ nhé
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\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)
\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)
\(P>0\Leftrightarrow x>-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
12 + 3 = 15 13 + 6 = 19 12 + 1 = 13
14 + 4 = 18 12 + 2 = 14 16 + 2 = 18
13 + 0 = 13 10 + 5 = 15 15 + 0 = 15
12 + 3 = 15 13 + 6 = 19 12 + 1 = 13
14 + 4 = 18 12 + 2 = 14 16 + 2 = 18
13 + 0 = 13 10 + 5 = 15 15 + 0 = 15
Sai đề
ko khi nào mà x*6=13