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\(c)16^{20}\)và \(32^{15}\)
Ta có: \(16^{20}=\left(2^4\right)^{20}=2^{80}\)
\(32^{15}=\left(2^5\right)15=2^{75}\)
Vì \(2^{80}>2^{75}\)
\(\Rightarrow16^{20}>32^{15}\)
Vậy \(16^{20}>32^{15}\)
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)
Ta lấy vễ trên chia vế dưới
\(=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)
Ta lấy vế trên chia vế dưới
\(=2^3.3=24\)
\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)
con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
Ta có:
200920 = 200910.200910 < 200910.1000110 = 2009200910
=> 200920 < 2009200910
Từ \(2y^3-1=15\Rightarrow2y^3=16\Rightarrow y^3=8=2^3\Rightarrow y=2\)
Thay y = 2 vao \(x+\frac{16}{9}=y+\frac{30}{16}\)duoc :
\(x+\frac{16}{9}=2+\frac{15}{8}=\frac{31}{8}\Rightarrow x=\frac{31}{8}-\frac{16}{9}=\frac{151}{72}\)
huhu trả lời dùm mik đi mai mik phải nộp rồi :'9