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a, A = \(\dfrac{3^9\times3^{20}\times2^8}{3^{24}\times243\times2^6}\)
A = \(\dfrac{3^{29}\times2^8}{3^{24}\times3^5\times2^6}\)
A = \(\dfrac{3^{39}\times3^{20}\times2^8}{3^{29}\times2^6}\)
A = 22
A = 4
b, \(\dfrac{2^{15}\times5^3\times2^6\times3^4}{8\times2^{18}\times81\times5}\)
B = \(\dfrac{2^{21}\times3^4\times5^3}{2^3\times2^{18}\times3^4\times5}\)
B = \(\dfrac{2^{21}\times3^4\times5^3}{2^{21}\times3^4\times5}\)
B = 52
B = 25
2: \(=\dfrac{-2}{75}+\dfrac{5}{39}=\dfrac{33}{325}\)
3: \(=\dfrac{6}{11}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)=\dfrac{6}{11}\)
4: \(=\dfrac{7}{19}\left(\dfrac{5}{13}+\dfrac{8}{13}-1\right)=-2\cdot\dfrac{7}{19}=-\dfrac{14}{19}\)
5: \(=\dfrac{2}{7}\left(\dfrac{4}{23}-\dfrac{27}{23}+1\right)=0\)
6: \(=\dfrac{3}{8}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\dfrac{11}{8}=\dfrac{3}{8}+\dfrac{11}{8}=\dfrac{14}{8}=\dfrac{7}{4}\)
a) Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\)
=> (3 + x).5 = 3(5 + y)
=> 15 + 5x = 15 + 3y
=> 5x = 3y
=> x = 3/5y
Mà x + y = 16
hay 3/5y + y = 16
=> (3/5 + 1).y = 16
=> 8/5.y = 16
=> y = 16 : 8/5
=> y = 10
=> x = 16 - 10 = 6
Vậy x = 6; y = 10
b) Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\)
=> (x - 7).6 = 7.(y - 6)
=> 6x - 42 = 7y - 42
=> 6x = 7y
=> x = 7/6y
Mà x - y = -4
hay 7/6y - y = -4
=> 1/6y = -4
=> y = -4 : 1/6
=> y = -24
=> x = -4 - 24 = -28
Vậy x = -28; y = -24
\(a,\Rightarrow\left(35x+3\right)\cdot19=152\\ \Rightarrow35x+3=8\\ \Rightarrow x=\dfrac{1}{7}\\ b,\Rightarrow3\left(x+7\right)=42\\ \Rightarrow x+7=14\Rightarrow x=7\\ c,\Rightarrow3\left(x+1\right)=48\\ \Rightarrow x+1=16\Rightarrow x=15\\ d,\Rightarrow120-5x+100\cdot2:5=4\cdot15\\ \Rightarrow120-5x+40=60\\ \Rightarrow5x=100\Rightarrow x=20\\ e,\Rightarrow4x-10=30\\ \Rightarrow4x=40\\ \Rightarrow x=10\\ g,\Rightarrow10x+10=70\\ \Rightarrow10x=60\\ \Rightarrow x=6\)
\(x=\frac{-1}{2}+\frac{3}{4}=\frac{-2}{4}+\frac{3}{4}=\frac{1}{4}\)
\(\frac{x}{5}=\frac{5}{6}-\frac{19}{30}=\frac{25}{30}-\frac{19}{30}=\frac{6}{30}=\frac{1}{5}\)
*\(1\div2\)không phải \(\frac{1}{2}\)à?*