hơi dài á

a, ta có:

kẻ đường cao CH

=> S_ABC \(=\frac{CHxAB}{2}\) ; S_CDB \(=\frac{BDxCH}{2}\)

mà BD= \(\frac{1}{4}\)AB nên S_CDB=\(\frac{1}{4}\)xS_ABC

CMTT với S_BEC = \(\frac{1}{4}\)xS_ABC

_{b, ta có:} S_CDB = S_BEC =  \(\frac{1}{4}\)xS_ABC

_=> S_DBG+S_BGC = S_BGC+S_EGC

_=> S_DBG=S_EGC

ccccgggggggggg

a: \(\frac{375\cdot833+822}{377\cdot833-844}\)

\(=\frac{375\cdot833+822}{375\cdot833+2\cdot833-844}=\frac{375\cdot833+822}{375\cdot833+822}\)

=1

b: \(\frac{134^2-670}{133^2-532}=\frac{134\left(134-5\right)}{133\left(133-4\right)}=\frac{134\cdot129}{133\cdot129}=\frac{134}{133}=1+\frac{1}{133}\)

\(\frac{7\cdot\left(13\cdot14-26\cdot28\right)}{26\cdot28-52\cdot56}=\frac{7\cdot13\cdot14\cdot\left(1-2\cdot2\right)}{26\cdot28\left(1-2\cdot2\right)}\)

\(=\frac{7\cdot13\cdot14}{26\cdot28}=7\cdot\frac12\cdot\frac12=\frac74=1+\frac34\)

Ta có: \(\frac{1}{133}=\frac{3}{399}<\frac34\)

=>\(\frac{1}{133}+1<\frac34+1\)

=>\(\frac{134^2-670}{133^2-532}<\frac{7\cdot\left(13\cdot14-26\cdot28\right)}{26\cdot28-52\cdot56}\)

c: \(\frac{2019\cdot2020-1}{2019\cdot2020}=1-\frac{1}{2019\cdot2020}\)

\(\frac{2020\cdot2021-1}{2020\cdot2021}=1-\frac{1}{2020\cdot2021}\)

Ta có: \(2019\cdot2020<2020\cdot2021\)

=>\(\frac{1}{2019\cdot2020}>\frac{1}{2020\cdot2021}\)

=>\(-\frac{1}{2019\cdot2020}<-\frac{1}{2020\cdot2021}\)

=>\(-\frac{1}{2019\cdot2020}+1<-\frac{1}{2020\cdot2021}+1\)

=>\(\frac{2019\cdot2020-1}{2019\cdot2020}<\frac{2020\cdot2021-1}{2020\cdot2021}\)

d: \(31^8<32^8=\left(2^5\right)^8=2^{40}\)

\(128^6=\left(2^7\right)^6=2^{42}>2^{40}\)

=>\(31^8<128^6\)

=>\(\frac{1}{31^8}>\frac{1}{128^6}\)

=>\(\left(\frac{1}{31}\right)^8>\left(\frac{1}{128}\right)^6\)

ý bạn là gì sao vô đây nói không không vậy bạn ,có gì bạn nói nguyên đề bài đi.Mình giúp được thì mình giúp nhé

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