a)

Theo tính chất kề bù có:

\(\widehat{xOy}+\widehat{xOy'}=180^o\\ \Rightarrow\widehat{xOy'}=180^o-130^o=50^o\)

b)

Vì Ot là tia phân giác của \(\widehat{xOy}\) nên:

\(\widehat{tOx}=\dfrac{\widehat{xOy}}{2}=\dfrac{130^o}{2}=65^o\)

Vì Ot' là tia phân giác của \(\widehat{xOy'}\) nên:

\(\widehat{xOt'}=\dfrac{\widehat{xOy'}}{2}=\dfrac{50^o}{2}=25^o\)

Số đo góc \(\widehat{tOt'}\) là:

\(\widehat{tOt'}=\widehat{tOx}+\widehat{xOt'}=65^o+25^o=90^o\)

a) 50^o

b) 90^o

bài 1:

\(1)x+\frac34=\frac43\Rightarrow x=\frac43-\frac34=\frac{7}{12}\)

\(2)x-\frac35=\frac{4}{15}\Rightarrow x=\frac{4}{15}+\frac35=\frac{13}{15}\)

\(3)-x-\frac{2}{15}=\frac{7}{10}\Rightarrow x=-\frac{2}{15}-\frac{7}{10}=-\frac56\)

\(4)\frac79-x=\frac{1}{12}\Rightarrow x=\frac79-\frac{1}{12}=\frac{25}{36}\)

\(5)\frac34x=\frac{-5}{6}\Rightarrow x=-\frac56:\frac34=-\frac56\cdot\frac43=-\frac{10}{9}\)

\(6)\left(x+\frac23\right)\cdot\frac45=-\frac{8}{15}\Rightarrow x+\frac23=-\frac{8}{15}:\frac45\)

\(x+\frac23=-\frac{8}{15}\cdot\frac54=-\frac23\)

\(x=-\frac23-\frac23=-\frac43\)

\(7)\frac67:x=-\frac{8}{21}\Rightarrow x=\frac67:\left(-\frac{8}{21}\right)\)

\(x=\frac67\cdot\left(-\frac{21}{8}\right)=-\frac94\)

\(8)x:\frac35=-\frac52\Rightarrow x=-\frac52\cdot\frac35=-\frac32\)

\(9)\frac34:\left(\frac14+x\right)=-\frac56\Rightarrow\frac14+x=\frac34:\left(-\frac56\right)\)

\(\frac14+x=\frac34\cdot\left(-\frac65\right)=-\frac{9}{10}\)

\(\Rightarrow x=-\frac{9}{10}-\frac14=-\frac{23}{20}\)

bài 2:

\(1)\frac34+\frac14x=-\frac56\Rightarrow\frac14x=-\frac56-\frac34\)

\(\frac14x=-\frac{19}{12}\Rightarrow x=-\frac{19}{12}:\frac14\)

\(x=-\frac{19}{12}\cdot4=-\frac{19}{3}\)

\(2)3\frac12-\frac12x=-\frac23\Rightarrow\frac72-\frac12x=-\frac23\)

\(\frac12x=\frac72-\left(-\frac23\right)=\frac{25}{6}\)

\(x=\frac{25}{6}:\frac12=\frac{25}{6}\cdot2=\frac{25}{3}\)

\(3)x:\frac12+\frac32=\frac54\Rightarrow x\cdot2=\frac54-\frac32\)

\(x\cdot2=-\frac14\Rightarrow x=-\frac14:2=-\frac14\cdot\frac12=-\frac18\)

\(4)\frac58-\frac38:x=\frac12\Rightarrow\frac38\cdot\frac{1}{x}=\frac58-\frac12\)

\(\frac{3}{8x}=\frac18\Rightarrow24=8x\Rightarrow x=3\)

\(5)\frac32x-\frac14=\frac47\Rightarrow\frac32x=\frac47+\frac14\)

\(\frac32x=\frac{23}{28}\Rightarrow x=\frac{23}{28}:\frac32=\frac{23}{28}\cdot\frac23=\frac{23}{42}\)

\(6)\frac13+\frac12:x=\frac15\Rightarrow\frac12\cdot\frac{1}{x}=\frac15-\frac13\)

\(\frac{1}{2x}=-\frac{2}{15}\Longrightarrow4x=-15\Rightarrow x=-\frac{15}{4}\)

Câu 5:

b) Xét ΔMNE vuông tại M và ΔDNE vuông tại D có 

NE chung

NM=ND(gt)

Do đó: ΔMNE=ΔDNE(cạnh huyền-cạnh góc vuông)

Bài này quá dễ 

x/40 - x/50= 1/3

<=>5x/200 - 4x/200=1/3

<=> x/200= 1/3

<=> x= 200/3.

bạn ê, mik bị ngu toán, lười suy nghĩ ấy mà nên đừng nói dễ hay khó j vs mik

Ờm, k cho mình đi

Bài 7: 

\(\widehat{C}=180^0-70^0-40^0=70^0=\widehat{B}\)