Do \(\widehat{xOz}\)và\(\widehat{yOz}\)là hai góc kề bù.

=> \(\widehat{xOz}+\widehat{yOz}=180^o\)

=>\(\widehat{yOz}=130^o\)

Có \(Ot\)là phân giác \(\widehat{yOz}\)

=> \(\widehat{zOt}=\widehat{tOy}=\frac{1}{2}\widehat{yOz}=\frac{1}{2}\times130^o=65^o\)

Lại có \(\widehat{xOt}=\widehat{xOz}+\widehat{zOt}=50^o+65^o=115^o\)

Vậy \(\widehat{xOt}=115^o\).

\(a.\left|-2x+1,5\right|=\frac{1}{4}\)

\(\Rightarrow\)\(-2x+1,5=\frac{1}{4}\)hoặc \(-\frac{1}{4}\)

\(\Rightarrow\)\(-2x=-\frac{5}{4}\)hoặc \(-\frac{7}{4}\)

\(\Rightarrow\)\(x=\frac{5}{8}\)hoặc \(\frac{7}{8}\)

Vậy x \(\in\){ ..... }

\(b.\frac{3}{2}-\left|1\frac{1}{4}+3x\right|=\frac{1}{4}\)

\(\left|\frac{5}{4}+3x\right|=\frac{3}{2}-\frac{1}{4}\)

\(\left|\frac{5}{4}+3x\right|=\frac{5}{4}\)

\(\Rightarrow\)\(\frac{5}{4}+3x=\frac{5}{4}\)hoặc \(-\frac{5}{4}\)

\(\Rightarrow\)\(3x=0\)hoặc \(\frac{-5}{2}\)

\(\Rightarrow\)\(x=0\)hoặc \(\frac{-5}{6}\)

Vậy x \(\in\){ ...... }

a)Ta có :\(\left|x+6\right|+\left|4-x\right|\ge\left|x+6+4-x\right|=\left|10\right|=10\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+6\right)\left(4-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x+6\ge0\\4-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+6\le0\\4-x\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-6\\x\le4\end{cases}}\)hoặc \(\hept{\begin{cases}x\le-6\\x\ge4\end{cases}}\)(Vô lí)

\(\Leftrightarrow-6\le x\le4\)

Vậy \(-6\le x\le4\)

b)Ta có :\(\left|x-1\right|+\left|x-4\right|=\left|x-1\right|+\left|4-x\right|\ge\left|x-1+4-x\right|=\left|3\right|=3\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-1\right)\left(x-4\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-1\ge0\\x-4\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1\le0\\x-4\le0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ge4\end{cases}}\)hoặc \(\hept{\begin{cases}x\le1\\x\le4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\ge4\\x\le1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x\ge4\\x\le1\end{cases}}\)

Nếu x < 1 => |-x + 1| = -x + 1

|2x - 3| = -(2x - 3) = -2x + 3

Khi đó B = |-x + 1| + |2x - 3| - 2(x - 1)

= -x + 1 - 2x + 3 - 2x + 2

= - 5x + 6

Nếu \(1\le x\le1,5\)

=> |-x + 1| = x - 1 

 |2x - 3| = --2x + 3

Khi đó B = x - 1 - 2x + 3 - 2x + 2

= -4x + 4

Nếu x > 1,5 => |-x + 1| = x - 1

|2x - 3| = 2x - 3

Khi đó B = x - 1 + 2x - 3 -2x + 2 

= x - 2 

how to đăng bài?

please help me