Bài 1:Ta có: \(\widehat{xAm}+\widehat{yAn}=130^0\)

mà \(\widehat{xAm}=\widehat{yAn}\)(hai góc đối đỉnh)

nên \(\widehat{xAm}=\widehat{yAn}=\dfrac{130^0}{2}=65^0\)

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

=>\(\widehat{xAn}+65^0=180^0\)

=>\(\widehat{xAn}=115^0\)

=>\(\widehat{yAm}=115^0\)
Bài 2:

Ta có: \(\widehat{xOz}+\widehat{zOt}+\widehat{tOy}=220^0\)

=>\(\widehat{xOz}+\widehat{tOy}=40^0\)

mà \(\widehat{xOz}=\widehat{tOy}\)(hai góc đối đỉnh)

nên \(\widehat{xOz}=\widehat{tOy}=\dfrac{40^0}{2}=20^0\)

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

=>\(\widehat{xOt}=180^0-20^0=160^0\)

=>\(\widehat{yOz}=160^0\)

\(a,-0,25+\dfrac{2}{3}=-\dfrac{3}{4}+\dfrac{2}{3}=-\dfrac{9}{12}+\dfrac{8}{12}=-\dfrac{1}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(-\dfrac{5}{21}-\dfrac{16}{21}\right)+0,5\\ =\dfrac{23}{23}-\dfrac{21}{21}+0,5\\ =1-1+0,5\\ =0,5\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-\dfrac{13}{9}-1\right]\\ =2-\left[\dfrac{4}{9}-\dfrac{13}{9}-\dfrac{9}{9}\right]\\ =2-\left(-2\right)\\ =4\)

\(a,-0,25+\dfrac{2}{3}\\ =-\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{-3}{12}+\dfrac{8}{12}\\ =\dfrac{5}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =1+\left(\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{-5}{21}-\dfrac{16}{21}\right)+\dfrac{1}{2}\\ =1+\dfrac{-21}{21}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\ =\dfrac{1}{2}\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1-\dfrac{4}{9}-1\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^2-2-\dfrac{4}{9}\right]\\ =2-\left(\dfrac{4}{9}-2-\dfrac{4}{9}\right)\\ =2+2\\ =4\)

Ta có:

`(x+2)^2>=0` với mọi x

`|2y-3|>=0` với mọi y

`=>A=(x+2)^2+|2y-3|+2024>=2024` với mọi x,y

Dấu "=" xảy ra: 

`x+2=0` và `2y-3=0`

`<=>x=-2` và `2y=3`

`<=>x=-2` và y=3/2`

Bài 8:

a) Ta có:

\(\widehat{N_1}+\widehat{N_2}=180^o\\ =>\widehat{N_1}=180^o-\widehat{N_2}=180^o-125^o=55^o\)

\(\widehat{M_1}=\widehat{N_1}=55^o\)

Mà hai góc này ở vị trí đồng vị 

`=>x`//`y`

b) Ta có:

\(\widehat{P_1}+\widehat{P_2}=180^o\\ =>\widehat{P_1}=180^o-\widehat{P_2}=180^o-140^o=40^o\)

\(\widehat{P_1}=\widehat{Q_1}=40^o\)

Mà hai góc này ở vị trí đồng vị 

`=>a`//`b` 

bài 1:

a: 

\(\dfrac{15}{8}=1,875;-\dfrac{99}{20}=-4,95;\dfrac{40}{9}=4,\left(4\right);-\dfrac{44}{7}=-6,\left(285714\right)\)

b: Các số thập phân vô hạn tuần hoàn là:

4,(4); (-6,285714)

Bài 7: Độ dài đường chéo hình vuông là:

\(\sqrt{5^2+5^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\left(cm\right)\)

Bài 6: Diện tích sân là:

\(10125000:125000=81\left(m^2\right)\)

Chiều dài cạnh của sân là: \(\sqrt{81}=9\left(m\right)\)

`(8-9/4 +2/7)-(6 -3/7 +5/4)-(3+ 2/4 -9/7)`

`= 8-9/4 +2/7-6 +3/7 -5/4 -3- 2/4 +9/7`

`= (8-6-3)-(9/4+5/4 + 2/4) +(2/7 +3/7   +9/7)`

`= -1 - 16/4 + 14/7`

`= -1 -4 + 2`

`= -3`

a: Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAC}+60^0+30^0=180^0\)

=>\(\widehat{BAC}=90^0\)

AD là phân giác của góc BAC

=>\(\widehat{BAD}=\widehat{CAD}=\dfrac{\widehat{BAC}}{2}=45^0\)

Ta có: \(\widehat{BAH}+\widehat{B}=90^0\)(ΔBHA vuông tại H)

=>\(\widehat{BAH}=90^0-60^0=30^0\)

Vì \(\widehat{BAH}< \widehat{BAD}\)

nên tia AH nằm giữa hai tia AB và AD

=>\(\widehat{BAH}+\widehat{HAD}=\widehat{BAD}\)

=>\(\widehat{HAD}=45^0-30^0=15^0\)

ΔAHD vuông tại H
=>\(\widehat{HAD}+\widehat{HDA}=90^0\)

=>\(\widehat{HDA}=90^0-15^0=75^0\)

Sửa đề: `S = 1/(2^2) - 1/(2^4) + 1/(2^6) - ... - 1/(2^2020) `

`=> 2^2 S = 1 - 1/(2^2) + 1/(2^4) - ... - 1/(2^2018) `

`=> 4S + S = (1 - 1/(2^2) + 1/(2^4) - ... - 1/(2^2018) ) + ( 1/(2^2) - 1/(2^4) + 1/(2^6) - ... - 1/(2^2020) )`

`=> 5S = 1 - 1/(2^2020) < 1`

`=> S  < 1/5 `

`=> S < 0,2 (đpcm)`

a; |6\(x\) + 22| + (y - 21)² = 0

    |6\(x+22\) | ≥ 0; (y - 21)² ≥ 0

    |6\(x\) + 22| + (y - 21)² = 0 ⇔ \(\left\{{}\begin{matrix}6x+22=0\\y-21=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}6x=-22\\y=21\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-\dfrac{22}{6}\\y=21\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-\dfrac{11}{3}\\y=21\end{matrix}\right.\)

Vậy (\(x\); y) = (- \(\dfrac{11}{3}\); 21)

b; 

A = |\(\dfrac{4}{3}\) - \(\dfrac{1}{4}\)| - \(\dfrac{2}{11}\)

A = |\(\dfrac{16}{12}\) - \(\dfrac{3}{12}\)| - \(\dfrac{2}{11}\)

A = | \(\dfrac{13}{12}\)| - \(\dfrac{2}{11}\)

A = \(\dfrac{13}{12}\) - \(\dfrac{2}{11}\)

A = \(\dfrac{143}{132}\)  - \(\dfrac{24}{132}\)

A = \(\dfrac{119}{132}\)

ĐKXĐ: x<>-1

\(C=\dfrac{x^2-1}{x+1}=\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{x+1}=x-1\)

=>Khi \(x\in Z\backslash\left\{-1\right\}\) thì C là số nguyên