\(\dfrac{6^5.\left(-12\right)^6}{\left(-4\right)^9.\left(-3\right)^{10}}\)
Mọi người làm chi tiết từng bước một giùm em nhé và em cần gấp
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\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)
\(=\left(\frac{3}{7}\right)^9\)
\(\dfrac{6^5.\left(-12\right)^6}{\left(-4\right)^9\left(-3\right)^{10}}\)
\(=\dfrac{6^5.12^6}{\left(-4\right)^9.3^{10}}\)
\(=\dfrac{2^5.3^5.2^{12}.3^6}{\left(-1\right).2^{18}.3^{10}}\)
\(=\dfrac{2^{17}.3^{11}}{\left(-1\right).2^{18}.3^{10}}\)
\(=\dfrac{3}{\left(-1\right).2}\)
\(=\dfrac{-3}{2}\)
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\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)
bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với
Ta có: \(\dfrac{6^5\cdot\left(-12\right)^6}{\left(-4\right)^9\cdot\left(-3\right)^{10}}\)
\(=-\dfrac{3^5\cdot2^5\cdot12^6}{4^9\cdot3^{10}}\)
\(=-\dfrac{2^5\cdot3^6\cdot4^6}{4^9\cdot3^5}\)
\(=-\dfrac{2^5\cdot3}{4^3}\)
\(=-\dfrac{2^5}{2^6}\cdot3=-\dfrac{3}{2}\)
ta được \(\dfrac{6^5.12^6}{4^8.\left(-4\right).3^{10}}\) \(=\dfrac{2^5.3^5.2^{12}.3^6}{2^{16}.\left(-4\right).3^{10}}\) \(=\dfrac{2^{17}.3^{11}}{2^{16}.\left(-4\right).3^{10}}=\dfrac{-6}{4}=\dfrac{-3}{2}\)