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x+13=15
x = 15 - 13
x = 2
=> Vậy x = 2
~ Ủng HỘ Mk Nha ~
\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)
\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)
\(P>0\Leftrightarrow x>-4\)
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
\(\frac{x}{1.3}+\frac{x}{3.5}+\frac{x}{5.7}+.....+\frac{x}{9.11}+\frac{x}{11.13}=\frac{6}{13}\)
=>\(\frac{x}{1}-\frac{x}{3}+\frac{x}{3}-\frac{x}{5}+\frac{x}{5}-\frac{x}{7}+......+\frac{x}{9}-\frac{x}{11}+\frac{x}{11}-\frac{x}{13}=\frac{12}{13}\)
\(\frac{x}{1}-\frac{x}{13}=\frac{12}{13}\)
x.12/13=12/13 => x=1
15 + 1 = 16 10 + 2 = 12 14 + 3 = 17 13 + 5 = 18
18 + 1 = 19 12 + 0 = 12 13 + 4 = 17 15 + 3 = 18
15+1=16 10+2=12 14+3=17 13+5=18 18+1=19 12+0=12 13+4=17 15+3=18
x=25
x=13 hoặc 0
x= 1 hoặc 0
a) ( x - 25 ). 13 = 0
=> x - 25 = 0
=> x = 0 + 25
=> x = 25
b) x (x - 13) = 0
\(\Rightarrow\hept{\begin{cases}x=0\\x-13=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=13\end{cases}}}\)
c) (x-1).(3x-15)=0
\(\Rightarrow\hept{\begin{cases}x-1=0\\3x-15=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=5\end{cases}}}\)