a/9=b/11 và a+b=20
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\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)
\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)
\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)
\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)
Dễ thấy \(A=\frac{147}{10}>1\)
Mà \(B=\frac{43}{48}< 1\)
=> tự so sánh
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a) \(\left(-9\right)+\left(-5\right)< \left(-7\right)\)
b) \(\left(-20\right)+14< 0\)
c) \(\left(-10\right)+23< 30\)
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b: 99^20=(99^2)^10=9801^10
=>99^20<9999^10
d: 10^10=100^5=4*50^5<48*50^5
e: 1990^10+1990^9
=1990^9(1990+1)
=1990^9*1991
1991^10=1991^9*1991
=>1991^10>1990^9*1991
=>1991^10>1990^10+1990^9
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a: =-1+17/20=-3/20
b: =(28/60-33/60)*(-25/3)
=(-1/12)*(-25/3)=1/12*25/3=25/36
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}=\dfrac{1}{9}\)
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a: \(33^{44}=\left(33^4\right)^{11}\)
\(44^{33}=\left(44^3\right)^{11}\)
mà \(33^4>44^3\)
nên \(33^{44}>44^{33}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{a}{9}=\frac{b}{11}=\frac{a+b}{9+11}=\frac{20}{20}=1\)
\(\Rightarrow\hept{\begin{cases}a=9\\b=11\end{cases}}\)