Do \(\widehat{xOy}\) và \(\widehat{yOz}\) là 1 góc kề bù nên:

\(\widehat{xOy}+\widehat{yOz}=180^0\)

\(\Rightarrow2.\widehat{yOz}+\widehat{yOz}=180^0\) (do \(\widehat{xOy}=2.\widehat{yOz}\))

\(\Rightarrow3.\widehat{yOz}=180^0\)

\(\Rightarrow\widehat{yOz}=180^0:3=60^0\)

\(\Rightarrow\widehat{xOy}=2.\widehat{yOz}=2.60^0=120^0\)

Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOz}=180^0-50^0=130^0\)

Ot là phân giác của góc xOz

=>\(\widehat{zOt}=\dfrac{\widehat{xOz}}{2}=65^0\)

Ta có: \(\widehat{zOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{yOt}=180^0-65^0=115^0\)

a: \(x=\left(x^3\right)^{\dfrac{1}{3}}\)

b: \(x=\left(x^5\right)^{\dfrac{1}{5}}\)

\(\dfrac{1}{3^6}=\dfrac{1}{3^4\cdot3^2}=\dfrac{1}{81\cdot9}=\dfrac{1}{729}\)

  \(\dfrac{1}{3^6}\) = \(\dfrac{1}{3^4.3^2}\) = \(\dfrac{1}{81.9}\) = \(\dfrac{1}{729}\) 

\(\left(-5\right)^5=\left(-5\right)^4\cdot\left(-5\right)=5^4\cdot\left(-5\right)=625\cdot\left(-5\right)=-3125\)

chịu

\(\left(x^2-1\right)\left(x^2+2\right)< 0\)

mà \(x^2+2>0\forall x\)

nên \(x^2-1< 0\)

=>\(x^2< 1\)

=>-1<x<1

Em kiểm tra lại đề, vế phải là \(\dfrac{x-1}{2023}+\dfrac{x}{2024}\) mới đúng

Ta có:
\(\dfrac{a}{b}< \dfrac{c}{d}\\ \Rightarrow ad< bc\\ \Rightarrow\left\{{}\begin{matrix}ad+ab< bc+ab\\ad+cd< bc+cd\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a\left(b+d\right)< b\left(a+c\right)\\d\left(a+c\right)< c\left(b+d\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{a+c}{b+d}\\\dfrac{c}{d}>\dfrac{a+c}{b+d}\end{matrix}\right.\\ \Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
Vậy...

Giải thích chi tiết một chút cho bạn dễ hiểu:
+)
 \(\dfrac{a}{b}< \dfrac{c}{d}\\ \Rightarrow\dfrac{a}{b}.bd< \dfrac{c}{d}.bd\\ \Rightarrow ad< bc\)

+) 
\(\left\{{}\begin{matrix}a\left(b+d\right)< b\left(a+c\right)\\d\left(a+c\right)< c\left(b+d\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a\left(b+d\right)}{b\left(b+d\right)}< \dfrac{b\left(a+c\right)}{b\left(b+d\right)}\\\dfrac{d\left(a+c\right)}{c\left(a+c\right)}< \dfrac{c\left(b+d\right)}{c\left(a+c\right)}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{a+c}{b+d}\\\dfrac{d}{c}< \dfrac{b+d}{a+c}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}< \dfrac{a+c}{b+d}\\\dfrac{c}{d}>\dfrac{a+c}{b+d}\end{matrix}\right. \)

A = \(\dfrac{4}{4}\) - 3|\(x-2\)|

A = 1 - 3|\(x-2\)|

Vì |\(x-2\)| ≥ 0 \(\forall\) \(x\) ⇒ 3.|\(x-2\)| ≥ 0 

Vậy 1 - 3|\(x-2\)| ≥ 1  dấu bằng xảy ra khi \(x-2\) = 0 ⇒ \(x=2\)

Vậy giá trị nhỏ nhất của A là 1 xảy ra khi \(x\) = 2