mọi người giúp mk với nhé

\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{5}{5}=1=B\)

a/

\(\frac{2001}{2004}=\frac{2004-3}{2004}=1-\frac{3}{2004}=1-\frac{1}{668}.\)

\(\frac{39}{40}=\frac{40-1}{40}=1-\frac{1}{40}\)

Ta có \(40< 668\Rightarrow\frac{1}{40}>\frac{1}{668}\Rightarrow1-\frac{1}{40}< 1-\frac{1}{668}\Rightarrow\frac{39}{40}< \frac{2001}{2004}\)

b/

\(A< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=1=B\)

\(\frac{-4}{3}=\frac{-28}{21};\frac{9}{-7}=\frac{-27}{21}\Rightarrow\frac{-4}{3}

ta có \(\frac{3}{7}=\frac{27}{63}\); \(\frac{4}{9}=\frac{28}{63}\)

vì \(\frac{27}{63}< \frac{28}{63}\)

nên\(\frac{3}{7}< \frac{4}{9}\)

\(8^5=\left(2^3\right)^5=2^{15}=2\times2^{14}\)

\(3\times4^7=3\times\left(2^2\right)^7=3\times2^{14}\)

vì 3>2 nên \(3\times2^{14}>2\times2^{14}\)hay \(3\times4^7>8^5\)

Ta có:

\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{36}}\)

Vì 2³⁵ < 2³⁶

=> \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)

=> \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

\(\left(\frac{1}{32}\right)^7=\left[\left(\frac{1}{2}\right)^5\right]^7=\left(\frac{1}{2}\right)^{35}\)và \(\left(\frac{1}{16}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^9=\left(\frac{1}{2}\right)^{32}\)

Mà:\(\left(\frac{1}{2}\right)^{35}>\left(\frac{1}{2}\right)^{32}\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

\(\left(\frac{1}{32}\right)^7=\left(\frac{1}{2}\right)^{35}\)

\(\left(\frac{1}{16}\right)^9=\left(\frac{1}{2}\right)^{32}\)

Do \(\left(\frac{1}{2}\right)^{35}>\left(\frac{1}{2}\right)^{32}\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

hi

ban de the ma ban ko la dc

\(a,\)\(\frac{12}{13}vs\frac{11}{14}\)

\(MSC=182\)

\(\frac{12}{13}=\frac{12.14}{13.14}=\frac{168}{182}\)

\(\frac{11}{14}=\frac{11.13}{14.13}=\frac{143}{182}\)

\(\frac{168}{182}>\frac{143}{182}\Rightarrow\frac{12}{13}>\frac{11}{14}\)

số hơi to !!

tóm lại là bạn quy đồng mẫu số  (  tính máy tính ) rồi so sánh.

Ta có : \(A-1=\frac{9^{11}+1}{9^{11}-7}-1=\frac{8}{9^{11}-7}\) ; \(B-1=\frac{9^{12}+3}{9^{12}-5}-1=\frac{8}{9^{12}-5}\)

Cần so sánh : \(9^{11}-7\) và \(9^{12}-5\)

Ta viết : \(9^{12}-5=9^{11}.9-5=9^{11}.\left(1+8\right)-5=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)\)

Xét : \(\left(9^{12}-5\right)-\left(9^{11}-7\right)=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)-\left(9^{11}-7\right)=8.9^{11}+2>0\)

\(\Rightarrow9^{12}-5>9^{11}-7\)

Do đó : \(B-1>A-1\Rightarrow B< A\)