Bài 5:

Ta có : \(\widehat{A_1}+\widehat{A_3}=180^o\) (kề bù)

            \(100^o+\widehat{A_3}=180^o\)

            \(\widehat{A_3}=80^o\)

Ta có: \(\widehat{A_3}=\widehat{B_1}=80^o\)

            \(\widehat{A_3}\) và \(\widehat{B_1}\) ở vị trí đồng vị 

\(\Rightarrow AC//BD\)

\(\Rightarrow\widehat{C}_1=\widehat{D_1}=135^o\) (đồng vị)

\(x=135^o\)

b)

Ta có: \(\widehat{G_1}+\widehat{B_1}=180^o\left(120^o+60^o=180^o\right)\)

               \(\widehat{G_1}\) và \(\widehat{B_1}\) ở vị trí trong cùng phía

\(\Rightarrow QH//BK\)

\(\Rightarrow\widehat{H_1}=\widehat{K_1}=90^o\)(so le)

\(x=90^o\)

b: \(\Leftrightarrow\left(2x-7\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

f: Ta có: \(\sqrt{2-x}=\sqrt{x^2-2x+4}\)

\(\Leftrightarrow x^2-2x+4=2-x\)

\(\Leftrightarrow x^2-x+2=0\)

\(\Leftrightarrow x\in\varnothing\)

21.C

22.D

23.C

24.A

1 C

2 B

3 C

4 A

5 D

II

1 A

2 B

3 A

4 C

5 B

6 B