\(\left|2x-\frac{3}{2}\right|-\frac{7}{4}=\frac{-3}{4}\)

\(\Leftrightarrow\left|2x-\frac{3}{2}\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{2}=1\\2x-\frac{3}{2}=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=\frac{1}{4}\end{cases}}\)

chịu bạn ạ mình khổng hiểu

Bài 2. 

1) \(\widehat{BIC}=180^o-\left(\widehat{IBC}+\widehat{ICB}\right)=180^o-\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=180^o-\frac{1}{2}.90^o=135^o\)

2) \(\widehat{BAH}=\widehat{BCA}\)(vì cùng phụ với góc \(\widehat{ABC}\))

\(\widehat{CAH}=\widehat{ABC}\)(vì cùng phụ với góc \(\widehat{ACB}\))

suy ra \(\widehat{BAM}=\widehat{BAH}+\widehat{HAM}=\widehat{ACB}+\frac{1}{2}\widehat{ABC}\)

mà \(\widehat{BMA}=\widehat{MAC}+\widehat{MCA}\)(góc ngoài tam giác)

\(=\frac{1}{2}\widehat{ABC}+\widehat{ACB}\)

suy ra \(\widehat{BAM}=\widehat{BMA}\)

suy ra đpcm.

3) \(\widehat{ABC}=180^o-\widehat{BAM}-\widehat{BMA}=180^o-2\widehat{AMB}=60^o\)

\(\Rightarrow\widehat{ACB}=90^o-\widehat{ABC}=30^o\)

Bài 1. 

1) \(AB//EF,CD\perp AB\Rightarrow CD\perp EF\).

2) Kẻ tia \(Cx\)song song với \(AB\).

Khi đó suy ra \(Cx//EF\).

Vì \(AB//Cx\Rightarrow\widehat{BAC}=\widehat{ACx}\)(hai góc so le trong) 

Vì \(EF//Cx\Rightarrow\widehat{CEF}=\widehat{ECx}\)(hai góc so le trong) 

suy ra \(\widehat{ACE}=\widehat{ACx}+\widehat{ECx}=\widehat{BAC}+\widehat{CEF}=25^o+35^o=60^o\)