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\(E=\dfrac{1}{31^7}\) \(F=\dfrac{1}{17^9}\)
ta có thể thấy tử đã tự rồi
giờ chỉ cần tìm mẫu của thằng nào bé hơn thì thằng đó lớn hơn
\(31^7ss17^9\)
ta có \(31^7< 32^7\Leftrightarrow31^7< 2^{5^7}=2^{35}\)
\(17^9>16^9\Leftrightarrow17^9>2^{36}\)
mà \(2^{36}>2^{35}\)
\(\Rightarrow17^9>31^7\)
\(\Leftrightarrow\dfrac{1}{17^9}< \dfrac{1}{31^7}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
1 )Ta có
\(M=\left(\dfrac{1}{2^2}-1\right)\cdot\left(\dfrac{1}{3^2}-1\right)\cdot\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right).....\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{3}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{4}{3}\cdot\dfrac{-3}{4}\cdot\dfrac{5}{4}\cdot\cdot\cdot\cdot\dfrac{-99}{100}\cdot\dfrac{101}{100}\)
\(=\dfrac{-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot3\cdot\left(-4\right)\cdot4\cdot\left(-5\right)\cdot5....\cdot\left(-100\right)\cdot100\cdot101}{2^2\cdot3^2\cdot4^2....\cdot100^2}\)
\(=-\dfrac{101}{200}< \dfrac{1}{2}\)
2 ) Số phân số của biểu thức B là 180 phân số
Ta có
\(\dfrac{1}{20}>\dfrac{1}{200};\dfrac{1}{21}>\dfrac{1}{200};\dfrac{1}{22}>\dfrac{1}{200};....;\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow B=\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}\cdot180=\dfrac{9}{10}\)
1. \(-\dfrac{52}{17}+\left(\dfrac{12}{19}+\dfrac{52}{17}\right)=\left[\left(-\dfrac{52}{17}\right)+\left(-\dfrac{52}{17}\right)\right]+\dfrac{12}{19}=\dfrac{12}{19}\)
2. \(\dfrac{21}{35}+\left(-1+\dfrac{14}{35}\right)=\dfrac{3}{5}+\left(-1+\dfrac{2}{5}\right)=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(-1\right)=1-1=0\)
1. \(\frac{-52}{17}+\left(\frac{12}{19}+\frac{52}{17}\right)=\frac{-52}{17}+\frac{12}{19}+\frac{52}{17}\)\(=\left(\frac{-52}{17}+\frac{52}{17}\right)+\frac{12}{19}=\frac{0}{17}+\frac{12}{19}=\frac{12}{19}\)
2. \(\frac{21}{35}+\left(-1+\frac{14}{35}\right)=\frac{21}{35}-1+\frac{14}{35}=\left(\frac{21}{35}+\frac{14}{35}\right)-1=1-1=0\)
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
Câu1:
a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)
\(=2008^2-\left(2008^2-4\right)\)
=4
b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)
c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)
=0
\(\left(\dfrac{1}{17}\right)^{14}=\dfrac{1}{17^{14}}< \dfrac{1}{16^{14}}=\dfrac{1}{\left(2^4\right)^{14}}=\dfrac{1}{2^{56}}\)
\(\left(\dfrac{1}{31}\right)^{11}=\dfrac{1}{31^{11}}>\dfrac{1}{32^{11}}=\dfrac{1}{\left(2^5\right)^{11}}=\dfrac{1}{2^{55}}\)
@Ribi Nkok Ngok