Mình không ghi lại đầu bài nhé bạn 

=\(\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)

=\(1+\left(-1\right)+\frac{-5}{7}\)

=\(\frac{-5}{7}\)

\(\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)

\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)

\(=1+\left(-1\right)+\frac{-5}{7}\)

\(=0+\frac{-5}{7}\)

\(=\frac{-5}{7}\)

CHÚC BN HỌC TỐT!!!!

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c>

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a) \(\left(x+\frac{1}{4}\right)^2+\frac{11}{25}=\frac{18}{25}\)

\(\Rightarrow\left(x+\frac{1}{4}\right)^2=\frac{7}{25}\)

\(\Rightarrow\) Không có x

\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)

\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)

\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)

\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)

Vậy A = \(\frac{-5}{7}\)

B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)

\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)

\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)

\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)

\(\Leftrightarrow B=0+\frac{-2}{11}\)

\(\Leftrightarrow\) \(B=\frac{-2}{11}\)

Vậy \(B=\frac{-2}{11}\)

@@ Học tốt

Chiyuki Fujito

K cần tk nhá

Thanks bn nha

a, -3/5=39/-65 vì (-3).(-65)=5.39=195

b, -9/27=-41/123 vì (-9).123=(-41).27=-1107

c, -3/4 \(\ne\) 4/-5 vì (-3).(-5)\(\ne\) 4.4 (15 \(\ne\) 16)

d, 2/-3 \(\ne\)-5/7 vì 2.7\(\ne\)(-3).(-5) (vì 14 \(\ne\)15)

a,-3/5=39/-65 vì (-3)×(-65)=5×39

b,-9/27=-41/123 vì (-9)×123=27×(-41)

c,-3/4 không bằng 4/-5 vì (-3)×(-5) không bằng 4×4

d,2/-3 không bằng -5/7 vì 2×7 không bằng (-3)×(-5)

Đặt S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)

Ta thấy S có 40 số hạng

ta có:

S=\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)=\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)

\(+\left(\frac{1}{71}+\frac{1}{72}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 100 số hạng)

>\(\left(\frac{1}{50}+...+\frac{1}{50}\right)+\left(\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{70}+...+\frac{1}{70}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)\)(mỗi 1 nhóm có 10 số hạng)

=\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\)=\(\frac{533}{840}\)>\(\frac{490}{840}\)=\(\frac{7}{12}\)

vậy S>\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{80}\)(đpcm)

a) \(\frac{31}{23}-\left(\frac{7}{23}+\frac{8}{23}\right)\)

\(=\frac{31}{23}-\frac{15}{23}\)

\(=\frac{16}{23}\)

b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)

\(=\frac{1}{3}-1+1\)

\(=\frac{1}{3}\)

c) \(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{52}-\frac{3}{11}\right)\)

\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{30}{45}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{2}{3}+\frac{17}{52}+\frac{3}{11}\)

\(=\frac{104+51}{156}+\frac{3}{11}\)

\(=\frac{155}{156}+\frac{3}{11}\)

\(=\frac{156}{156}-\frac{1}{156}+\frac{3}{11}\)

\(=1-\frac{1}{156}+\frac{3}{11}\)

\(=1-\left(\frac{11-468}{1716}\right)\)

\(=1-\frac{-457}{1716}\)

\(=1+\frac{457}{1716}\)

\(=\frac{2173}{1716}\)

a)31/23-(7/32+8/23)=31/23-7/32-8/23=(31/23-8/23)-7/32=1-7/32=25/32

\(a)\) Ta có : 

\(\frac{51}{85}=\frac{3}{5}\)

\(\frac{58}{145}=\frac{2}{5}\)

Vì \(\frac{3}{5}>\frac{2}{5}\) nên \(\frac{51}{85}>\frac{58}{145}\)

Vậy \(\frac{51}{85}>\frac{58}{145}\)

\(b)\) Ta có : 

\(\frac{69}{-230}=\frac{-3}{10}\)

\(\frac{-39}{143}=\frac{-3}{11}\)

Vì \(\frac{-3}{10}< \frac{-3}{11}\) nên \(\frac{69}{-230}< \frac{-39}{143}\)

Vậy \(\frac{69}{-230}< \frac{-39}{143}\)

\(c)\) Ta có : 

\(1+\frac{-7}{41}=\frac{34}{41}\)

\(1+\frac{13}{-47}=\frac{34}{47}\)

Vì \(\frac{34}{41}>\frac{34}{47}\) nên \(1+\frac{-7}{41}>1+\frac{13}{-47}\) hay \(\frac{-7}{41}>\frac{13}{-47}\)

Vậy \(\frac{-7}{41}>\frac{13}{-47}\)

\(d)\) Ta có : 

\(1-\frac{40}{49}=\frac{9}{49}\)

\(\frac{15}{21}=\frac{5}{7}=\frac{35}{49}< \frac{40}{49}\)

Vậy \(\frac{40}{49}>\frac{15}{21}\)

Giúp với :)) !!!!!

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