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\(=\frac{7^2.3^3-3^2.5^2.7+7.3^4}{3.3^2.5^2-2^3.5^3}\)\(=\frac{7.3^2\left(7.3-5^2+3^2\right)}{5^2\left(3^3-2^3.5\right)}\)\(=\frac{7.3^2.5}{5^2.\left(-13\right)}=\frac{7.3^2}{5.\left(-13\right)}=\frac{63}{-65}\)
\(a.\left(\frac{-2}{3}+\frac{4}{15}\right)^3=\left(\frac{-10+4}{15}\right)^3=\frac{-6^3}{15}=\frac{-8}{125}\)
\(b.\left(\frac{3}{21}-\frac{2}{7}\right)^2=\left(\frac{3}{21}-\frac{6}{21}\right)^2=\frac{-1}{7}^2=\frac{-1}{49}\)
\(d.3-\left(-3.15\right)^0+\left(0.5\right)^2:2=3-1+0.25=2+0.25=2.25\)
\(e.81-3^2:\left(0.375\right)^2=81-9:0.140625=81-64=17\)
- a) =\(\left(\frac{-10}{15}+\frac{4}{15}\right)^3\)=\(\left(\frac{-6}{15}\right)^3\) = \(\left(\frac{-2}{5}\right)^3\) =\(\frac{-8}{125}\)
b) \(=\left(\frac{3}{27}-\frac{6}{21}\right)^2=\left(\frac{-3}{21}\right)^2=\left(\frac{-1}{7}\right)^2=\frac{1}{49}\)
d) \(=3-1+1:2=2+\frac{1}{2}=\frac{5}{2}\)
e) \(=81-9:\left(\frac{3}{8}\right)^2=72:\frac{9}{64}=72.\frac{64}{9}=512\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
a/\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}.\left(\frac{49}{9}\right)^6\)
đơn giản rồi cộng số mũ nhá!
b/ \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2\)
\(=2-0,125\)
\(=1,875\)
k nha
Tìm x:
\(\dfrac{4-x}{6-x}\)=\(\dfrac{x-3}{x-8}\)\(\Rightarrow\)(4-x)(x-8)=(6-x)(x-3)
\(\Rightarrow\)12x-x2-32=9x-x2-18
\(\Rightarrow\)3x=14\(\Rightarrow\)x=\(\dfrac{14}{3}\).
\(\dfrac{49^{24}.125^{10}.2^8-5^{30}.7^{49}.4^5}{5^{29}.16^2.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8-5^{30}.7^{49}.2^{10}}{5^{29}.2^8.7^{48}}\)
=\(\dfrac{7^{48}.5^{30}.2^8.\left(1-7.2^2\right)}{5^{29}.2^8.7^{48}}\)
=5.(1-7.22) = 5.(1-28) = 5.(-27) = -135
a: \(=\dfrac{5^{42}\cdot7^{30}}{5^{40}\cdot7^{30}\cdot10}=\dfrac{5^2}{10}=\dfrac{25}{10}=\dfrac{5}{2}\)
b: \(=\dfrac{3^{18}\cdot5^{10}}{5^{10}\cdot3^{10}\cdot3^9}=\dfrac{1}{3}\)