Bài 4: 

Nhóm 1: x;1/3x; 8x

Nhóm 2: \(x^2;5x^2;-3x^2\)

lỗi ảnh r 

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Bài nào :>

lỗi r

a. \(\widehat{DAB}=\widehat{ABC}=\widehat{BCE}=90^0\)

\(\widehat{ABD}=180^0-\widehat{ABC}-\widehat{EBC}=180^0-60^0-\left(180^0-\widehat{BCE}-\widehat{CEB}\right)=180^0-60^0-\left(180^0-60-\widehat{CEB}\right)=\widehat{CEB}\)\(\Rightarrow\)△ABD∼△CEB (g-g).

\(\Rightarrow\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow AD.CE=CB.AB\Rightarrow AD.CE=a^2\) không đổi

b. \(\widehat{CAD}=\widehat{BAD}+\widehat{BAC}=60^0+60^0=\widehat{BCE}+\widehat{ACB}=\widehat{ACE}\)

 \(\dfrac{AD}{CB}=\dfrac{AB}{CE}\Rightarrow\dfrac{AD}{AC}=\dfrac{AC}{CE}\)

\(\Rightarrow\)△ACD∼△CEA (c-g-c) 

\(\Rightarrow\left\{{}\begin{matrix}\widehat{ACD}=\widehat{CEA}\\\dfrac{CE}{AC}=\dfrac{EA}{CD}\end{matrix}\right.\)

\(\Rightarrow\)△ACK∼△AEC (g-g).

\(\Rightarrow\dfrac{CK}{EC}=\dfrac{AK}{AC}\Rightarrow\dfrac{CE}{AC}=\dfrac{CK}{AK}\)

\(\Rightarrow\dfrac{AE}{CD}=\dfrac{CK}{AK}\Rightarrow AE.AK=CD.CK\)

Bài 4 nào bài 8 đâu em 

là sao

chắc lại bị lỗi rồi

mn xa lánh em quá hic:((((((

giúp em với mai em nộp bài đó ạ;-;

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b5 c làm ở dưới r í

bài 3.

1.\(\dfrac{-51}{136}=\dfrac{-3}{8};\dfrac{-60}{108}=\dfrac{-5}{9};\dfrac{26}{-156}=\dfrac{-1}{6}\)

\(\dfrac{-3}{8}=\dfrac{-24}{72};\dfrac{-5}{9}=\dfrac{-40}{72};\dfrac{-1}{6}=\dfrac{-12}{72}\)

2.\(\dfrac{-165}{270}=\dfrac{-11}{18};\dfrac{-91}{-156}=\dfrac{7}{12};\dfrac{-210}{1134}=\dfrac{-5}{27}\)

\(\dfrac{-11}{18}=\dfrac{-66}{108};\dfrac{7}{12}=\dfrac{63}{108};\dfrac{-5}{27}=\dfrac{-20}{108}\)

1. -51/136=-3/8; -60/108=-5/9;26/-156=-1/6

-3/8=-24/71;-5/9=-40/72;-1/6=-12/72