a: Ta có: \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{8}\\x=-\dfrac{3}{8}\end{matrix}\right.\)

b: Ta có: \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)

nên x=5

mn xa lánh em quá hic:((((((

giúp em với mai em nộp bài đó ạ;-;

refer:

b) Ta có \(\text{∠}yBC=360\text{ ° }-\left(\text{∠}yBA+\text{∠}ABC\right)=360\text{ ° }-\left(120\text{ ° }+90\text{ ° }\right)=150\text{ ° }\)

Mặt khác, ta có \(\text{∠}BCz=150\text{ ° }\)

\(\Rightarrow\text{∠}yBC=\text{∠}BCz\) mà hai góc này ở vị trí so le trong 

\(\Rightarrow By//Cz\) ( đpcm )

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 12: 

a: Xét ΔABM và ΔACN có

AB=AC

\(\widehat{ABM}=\widehat{ACN}\)

BM=CN

Do đó: ΔABM=ΔACN