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Toán 6 ?
\(\frac{4^3.9^5.\left(-2\right)^6}{16^4.\left(-27\right)^2}=\frac{\left(2^2\right)^3.\left(3^2\right)^5.2^6}{\left(2^4\right)^4.27^2}=\frac{2^6.3^{10}.2^6}{2^{16}.\left(3^3\right)^2}=\frac{2^{12}.3^{10}}{2^{16}.3^6}=\frac{3^4}{2^4}=\frac{81}{16}\)
\(\frac{4^3.9^5.\left(-2\right)^6}{16^4.\left(-27\right)^2}\)
= \(\frac{2^6.3^{10}.\left(-2\right)^6}{2^{16}.\left(-3\right)^6}\)
= \(\frac{2^{12}.3^{10}}{2^{16}.\left(-3\right)^6}\)
= \(\frac{3^4}{2^4}\)
= \(\left(\frac{3}{2}\right)^4\)
= \(\frac{81}{16}\)
Nếu bn thấy đúng k mk nhé! Cảm ơn nha
\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)
=>4x=18
hay x=9/2
2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)
=>4x=108
hay x=27
3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)
\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)
=>4x=12
hay x=3
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
a: \(=\dfrac{28-2-3}{4}:\dfrac{40-2-5}{8}=\dfrac{23}{4}\cdot\dfrac{8}{33}=\dfrac{46}{33}\)
b: =78(0,65+0,35)+2020(2,2-2,2)
=78*1=78
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right)+\dfrac{4}{5}\\ =-\dfrac{5}{21}:\dfrac{4}{5}+\dfrac{5}{21}\\ =\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\\ =0:\dfrac{4}{5}\\ =0.\)
Sửa cho mk dòng đầu là :4/5 và dòng tiếp theo mk thiếu :4/5
\(\dfrac{4^3\cdot9^5\cdot\left(-2\right)^6}{16^4\cdot\left(-27\right)^2}\)
=\(\dfrac{\left(2^2\right)^3\cdot\left(3^2\right)^5\cdot2^6}{\left(2^4\right)^4\cdot\left(-3^3\right)^2}\)
=\(\dfrac{2^6\cdot3^{10}\cdot2^6}{2^{16}\cdot\left(-3\right)^6}\)
=\(\dfrac{2^{6+6}\cdot3^{10}}{2^{16}\cdot3^6}\)
=\(\dfrac{2^{12}\cdot3^{10}}{2^{16}\cdot3^6}\)
=\(\dfrac{3^4}{2^4}\)
=\(\dfrac{81}{16}\)