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Câu 1:
a) \(-\dfrac{3}{7}-\left(\dfrac{2}{3}-\dfrac{3}{7}\right)=\dfrac{-3}{7}-\dfrac{2}{3}+\dfrac{3}{7}=\dfrac{-2}{3}\)
Câu 2:
b) \(\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{4}{5}-\dfrac{1}{3}\cdot\dfrac{6}{5}\right)=\dfrac{2}{15}:\left[\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{6}{5}\right)\right]=\dfrac{2}{15}:\left(\dfrac{1}{3}\cdot\dfrac{-2}{5}\right)=\dfrac{2}{15}:\dfrac{-2}{15}=\dfrac{2}{-2}=-1\)
B=1-2-3+4+5-6-7+8+..........+21-22-23+24
B=(1-2-3+4)+(5-6-7+8)+.......+(21-22-23+24)
B=0+0+............+0
B=0
a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)
b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)
1) Sửa đề :
\(\frac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\frac{10^3+5\cdot5^2\cdot2^2+5^3}{6^3+3\cdot2^2\cdot3^2+3^3}=\frac{5^3\cdot\left(2^3+2^2+1\right)}{3^3\cdot\left(2^3+2^2+1\right)}=\frac{5^3}{3^3}\)
2) Sửa đề :
\(\frac{5^3+3\cdot5^2}{-8}=\frac{5^2\cdot\left(5+3\right)}{-8}=\frac{-5^2\cdot8}{8}=-5^2=-25\)
1) \(=\frac{6^5.5^3\left(1+5\right)}{6^5.5^3.3}=\frac{6}{3}=2\)
2)
\(2B=2+2^2+2^3+...+2^{101}\)
\(2B-B=B=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)=2^{101}-1\)