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Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
d, ta có :(-32)9=-(329) ;(-18)13=-(1813)
329=32\(\times\)328=32\(\times\)(322)4=32\(\times\)10244=32\(\times\)1024\(\times\)10243
1813=18\(\times\)1812=18\(\times\)(183)4=18\(\times\)58324=18\(\times\)5832\(\times\)58323
18\(\times\)5832 >16\(\times\)5832=32\(\times\)2916>32\(\times\)1024 =58323>10243
nên 1813>329
vậy (-18)13 <(-32)9
(-32)9=-(329)
(-18)13=-(1813)
329<369
ta có :369=(2\(\times\)18)9=29\(\times\)189
vì 184>164mà 164=(24)4=216
mà 216>29
\(\Rightarrow\)184>29
\(\Rightarrow\)184\(\times\)189>29\(\times\)189
\(\Rightarrow\)1813>369mà 369 >329
\(\Rightarrow\)1813>329
\(\Rightarrow\)(-18)13<(-32)9
a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
Ta có :
\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{5.7}}=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{4.9}}=\frac{1}{2^{36}}\)
Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) ( cùng tử, mẫu nào bé hơn thì phân số đó lớn hơn ) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Vậy \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Chúc bạn học tốt ~
Ta có : \(\left(\frac{1}{32}\right)^7=\left(\frac{1}{2^5}\right)^7=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\left(\frac{1}{2^4}\right)^9=\frac{1}{2^{36}}\)
DO : \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)\(\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Tk mk nha !!!
Câu 1 :
a) \(4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)
\(=2^2.32^2:\left(\frac{1}{8}.16\right)=\left(2.32\right)^2:2=64^2:2\)
\(=2048=2^{11}\)
b) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)
\(=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)
VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ
\(a,4\cdot\left(\frac{1}{32}\right)^{-2}:\left(2^3\cdot\frac{1}{16}\right)\\ =4\cdot1024:\left(8\cdot\frac{1}{16}\right)\\ =4\cdot1024:\frac{1}{2}\\ =2\cdot1024\\ =2\cdot2^{10}\\ =2^{11}\)
\(b,5^2\cdot3^5\cdot\left(\frac{3}{5}\right)^2\\ =5^2\cdot\left(\frac{3}{5}\right)^2\cdot3^5\\ =3^2\cdot3^5\\ =3^7\)
2 SO SÁNH
\(a,10^{20}\text{ và }9^{10}\)
Có: \(9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(\Rightarrow10^{20}>3^{20}\\ \text{hay}\text{ }10^{20}>9^{10}\)
\(b,\left(-5\right)^3\text{ và }\left(-3\right)^{50}\)
Có: \(\left(-3\right)^{50}=3^{50}\)
\(\Rightarrow\left(-5\right)^3< 3^{50}\\ \text{hay }\left(-5\right)^3< \left(-3\right)^{50}\)
\(c,64^3\text{ và }16^{12}\)
Có: \(64^3=\left(4^3\right)^3=4^9;16^{12}=\left(4^2\right)^{12}=4^{24}\)
\(\Rightarrow4^9< 4^{24}\\ hay\text{ }64^3< 16^{12}\)
\(d,\left(\frac{1}{16}\right)^{10}\text{ và }\left(\frac{1}{2}\right)^{50}\)
Có: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\frac{1}{2}.\left(\frac{20}{15}+\frac{6}{15}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{48}{9}\right)\)
\(=\frac{1}{2}.\frac{26}{15}-\frac{3}{4}.\frac{56}{9}\)
\(=\frac{13}{15}-\frac{14}{3}\)
\(=-\frac{19}{5}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\left(\frac{1}{2}.\frac{4}{3}+\frac{1}{2}.\frac{2}{5}\right)-\left(\frac{3}{4}.\frac{8}{9}+\frac{3}{4}.\frac{16}{3}\right)\)
\(=\left(\frac{2}{3}+\frac{1}{5}\right)-\left(\frac{2}{3}+4\right)\)
\(=\frac{2}{3}+\frac{1}{5}-\frac{2}{3}-4\)
\(=\frac{1}{5}-4\)
\(=\frac{1}{5}-\frac{20}{5}=\frac{-19}{5}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)+\left(x+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow5x+\frac{31}{32}=1\frac{31}{32}\)
\(\Leftrightarrow5x=1\frac{31}{32}-\frac{31}{32}\Leftrightarrow5x=1\Rightarrow x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(\text{a) }\left(-\frac{1}{16}\right)^{100}=\frac{\left(-1\right)^{100}}{16^{100}}=\frac{1}{16^{100}}\)
\(\left(-\frac{1}{2}\right)^{500}=\frac{\left(-1\right)^{500}}{2^{500}}=\frac{1}{\left(2^5\right)^{100}}=\frac{1}{32^{100}}\)
Ta co
\(16^{100}< 32^{100}\)
\(\Rightarrow\frac{1}{16^{100}}>\frac{1}{32^{100}}\)
\(\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)
a.
Ta có:
\(\left(-\frac{1}{16}\right)^{100}=\frac{\left(-1\right)^{100}}{16^{100}}=\frac{1}{16^{100}}\)
\(\left(-\frac{1}{2}\right)^{500}=\frac{\left(-1\right)^{500}}{2^{500}}=\frac{1}{\left(2^5\right)^{100}}=\frac{1}{32^{100}}\)
Vì \(\frac{1}{16^{100}}>\frac{1}{32^{100}}\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)
b.
Ta có:
\(\left(-32\right)^9=\left[-\left(2^5\right)\right]^9=-\left(2^{45}\right)\)
\(\left(-16\right)^{13}=\left[-\left(2^4\right)\right]^{13}=-\left(2^{52}\right)\)
Vì \(-\left(2^{45}\right)>-\left(2^{52}\right)\Rightarrow\left(-32\right)^9>\left(-16\right)^{13}\)
#Chúc bạn học tốt!#