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\(-\dfrac{3}{7}\times\dfrac{15}{13}-\dfrac{3}{7}\times\dfrac{11}{13}-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\left(\dfrac{15}{13}+\dfrac{11}{13}+1\right)\)
\(=-\dfrac{3}{7}\left(\dfrac{15}{13}+\dfrac{11}{13}+\dfrac{13}{13}\right)\)
\(=-\dfrac{3}{7}\times\dfrac{39}{13}=-\dfrac{3}{7}\times3=-\dfrac{9}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(4-x\right)y+3x=15\)
\(\left(4-x\right)y+3x-15=0\)
\(-\left[\left(x-4\right)y-3x15\right]=0\)
\(\left(x-4\right)y-3x+15=0\)
\(x-4=0\)
\(x=4\)
\(y-3=0\)
\(y=3\)
\(=>x=4;y=3\)
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\(a,\frac{3}{20}=\frac{3\times3}{20\times3}=\frac{9}{60}\)
\(\frac{7}{30}=\frac{7\times2}{30\times2}=\frac{14}{60}\)
\(\frac{4}{15}=\frac{4\times4}{15\times4}=\frac{16}{60}\)
\(c,\frac{7}{-30}=\frac{7\times4}{\left(-30\right)\times4}=\frac{28}{-120}\)
\(\frac{3}{-40}=\frac{3\times3}{\left(-40\right)\times3}=\frac{9}{-120}\)
\(-\frac{11}{30}=\frac{\left(-11\right)\times4}{30\times4}=-\frac{44}{120}\)
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a: \(=\dfrac{-39+19+10}{12}=\dfrac{-10}{12}=\dfrac{-5}{6}\)
b: \(=\dfrac{2^{30}\cdot3^{16}\cdot7-2^{34}\cdot3^{15}}{2^{28}\cdot3^{21}-2^{28}\cdot3^{17}}\)
\(=\dfrac{2^{30}\cdot3^{15}\left(3\cdot7-2^4\right)}{2^{28}\cdot3^{17}\left(3^4-1\right)}=\dfrac{2^2}{3^2}\cdot\dfrac{21-16}{80}=\dfrac{4}{9}\cdot\dfrac{5}{80}\)
\(=\dfrac{20}{720}=\dfrac{1}{36}\)
4x3 + 15 = 47
4x3 = 32
x3 = 8
=> x = 2
4.2x - 3 = 125
4.2x = 128
2x = 32
=> x = 5
**** cho mình nha
\(a,4.x^3+15=47\)
\(4.x^3=47-15\)
\(4.x^3=32\)
\(x^3=32:4\)
\(x^3=8\)
\(x^3=2^3\)
\(=>x=2\)
Vậy \(x=2\)
\(b,4.2^x-3=125\)
\(4.2^x=125+3\)
\(4.2^x=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(=>x=5\)
Vậy \(x=5\)