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\(\frac{\frac{1}{3}+\frac{1}{9}-\frac{1}{27}}{\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}=\frac{\frac{9}{27}+\frac{3}{27}-\frac{1}{27}}{\frac{45}{27}+\frac{15}{27}-\frac{5}{27}}=\frac{\frac{9+3-1}{27}}{\frac{45+15-5}{27}}=\frac{\frac{11}{27}}{\frac{55}{27}}=\frac{11}{55}=\frac{1}{5}.\)
a) (−27). 1011 − 27. (−12) + 27. (−1)
= (-27). (1011 - 12 + 1)
= (-27). 1000 = -27 000
b) (−9). (−9). (−9) + 103 + 9.3
= (-9). 3 + 103 + 9.3
= 9.3 - 9.3 + 103
= 0 + 103 = 103
c) (−157). (127 − 316) − 127. (316 − 157)
= (-157). 127 - 316. (-157) - 127.316 - 127. (-157)
= (-157). 127 + 316. 157 - 127. 316 + 127.157
= (127.157 - 157.127) + (316. 157 - 127.316)
= 0 + (157 - 127). 316
= 0 + 30. 316
= 9480
a) (-27).1011-27.(-12)+27.(-1)
= -27.1011 +27.12+27.(-1)
= 27.(-1011)+27.12+27.(-1)
= 27.(-1011+12-1)
= 27.(-1000)
=-27000
c) (-157)(127-316)-127(316-157)
= (-157).127+157.316-127.316+127.127
= 316.(157-127)
= 316.30
= 9480
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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
\(1+\frac{9}{16}.\frac{13}{27}+\frac{9}{16}.\frac{4}{27}=1+\frac{9}{16}\left(\frac{13}{27}+\frac{4}{27}\right)=1+\frac{9}{16}.\frac{17}{27}=1+\frac{17}{48}=1\frac{17}{48}\)