So sánh : a) A= 10/2^7+ 10/2^6 và B= 11/2^7+9/2^6
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Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{5}{2^6}+\frac{10}{2^6}=\frac{15}{2^6}\)
Lại có B = \(\frac{11}{2^7}+\frac{9}{2^6}=\frac{5,5}{2^6}+\frac{9}{2^6}=\frac{14,5}{2^6}\)
Vì \(\frac{15}{2^6}>\frac{14,5}{2^6}\Rightarrow A>B\)
b) Ta có : \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\)
Lại có B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=\frac{-157}{10^{2006}}\)
Vì \(\frac{-85}{10^{2006}}>\frac{-157}{10^{2006}}\Rightarrow A< B\)
a: -3/4=-9/12
-5/6=-10/12
mà -9>-10
nên -3/4>-5/6
b: -5/17<0<2/7
c: 11/10>1>9/14
a) \(4\frac{7}{10}< 6\frac{7}{10}\)(4 < 6)
b) \(3\frac{4}{15}< 3\frac{11}{15}\)(4/15 < 11/15)
c) \(5\frac{1}{9}>2\frac{2}{5}\)(5 > 2)
d) \(2\frac{2}{3}=2\frac{10}{15}\)(10/15 = 2/3)
Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
A = \(-1+1+\frac{1}{2}\)
A = \(\frac{1}{2}\)
B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)
B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)
B = \(1+1-\frac{11}{27}\)
B = \(\frac{43}{27}\)
Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)
=> A < B
Giải
\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)
\(\Leftrightarrow A=-1+1+\frac{1}{2}\)
\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)
\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)
\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)
\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)
\(\Leftrightarrow B=1+\frac{-11}{27}+1\)
\(\Leftrightarrow B=2+\frac{-11}{27}\)
\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)
Từ (1) và (2) suy ra A < B
Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{10}{2^7}+\frac{20}{2^7}=\frac{30}{2^7}\)
Lại có \(B=\frac{11}{2^7}+\frac{9}{2^6}=\frac{11}{2^7}+\frac{18}{2^7}=\frac{29}{2^7}\)
Vì \(\frac{30}{2^7}>\frac{29}{2^7}\)
=> A > B
U