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: \(\sqrt{81}=9\left(m\right)\)

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`

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

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

\(\left|x-1\right|>=0\forall x;\left(x+y-2\right)^{2024}>=0\forall x,y\)

Do đó: \(\left|x-1\right|+\left(x+y-2\right)^{2024}>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\x+y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-x+2=-1+2=1\end{matrix}\right.\)

Thay x=1;y=1 vào Q, ta được:

\(Q=1^{2024}+1^{2024}=1+1=2\)

\(\left|x-1\right|+\left(x+y-2\right)^{2024}=0\)

Do \(\left|x-1\right|\ge0;\left(x+y-2\right)^{2024}\ge0,\forall x;y\in R\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\x+y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

\(Q=x^{2024}+y^{2024}=1^{2024}+1^{2024}=2\)

Lấy mẫu số, chứ không phải lấy tử số em nhé!

tất cả các số âm hay dương gì đều lấy mẫu số ạ?

Độ dài cạnh huyền là:

\(\sqrt{3^2+7^2}=\sqrt{9+49}=\sqrt{58}\left(cm\right)\)

Bình phương cạnh huyền là:

 3² + 7² =  58(cm²)

Cạnh huyền là:  \(\sqrt{58}\) m

1: \(\left(x^2+2xy-3\right)\left(-xy^2\right)\)

\(=-xy^2\cdot x^2-xy^2\cdot2xy+3\cdot xy^2\)

\(=-x^3y^2-2x^2y^3+3xy^2\)

2: \(3x\left(x+2\right)-3x^2-12=0\)

=>\(3x^2+6x-3x^2-12=0\)

=>6x-12=0

=>6x=12

=>x=2

3: \(\left(2x^3-\dfrac{9}{2}x^2+\dfrac{1}{xy}\right)\cdot x^2y^3\)

\(=2x^3\cdot x^2y^3-\dfrac{9}{2}x^2\cdot x^2y^3+\dfrac{x^2y^3}{xy}\)

\(=2x^5y^3-\dfrac{9}{2}x^4y^3+xy^2\)

2; 3\(x\)(\(x+2\)) - 3\(x^2\) - 12 = 0

    3\(x^2\) + 6\(x\) - 3\(x^2\) - 12 = 0

  (3\(x^2\) - 3\(x^2\)) + 6\(x\) - 12 = 0

   0 + 6\(x\) - 12 = 0

          6\(x\)  = 12

           \(x\) = 12 : 6

           \(x=2\) 

Vậy \(x=2\)

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)

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

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

Bài 1:

a: \(\dfrac{2}{3}-\dfrac{7}{6}+\dfrac{5}{2}=\dfrac{4}{6}-\dfrac{7}{6}+\dfrac{15}{6}=\dfrac{12}{6}=2\)

b: \(9-2023^0+\sqrt{\dfrac{1}{25}}=9-1+\dfrac{1}{5}=8+\dfrac{1}{5}=8,2\)

c: \(\dfrac{4^{1010}\cdot9^{1010}}{3^{2019}\cdot16^{504}}=\dfrac{4^{1010}}{4^{1008}}\cdot\dfrac{3^{2020}}{3^{2019}}=\dfrac{3}{4^8}\)

Bài 3:

Tổng số tiền phải trả cho 1 bánh cỡ to, 2 bánh cỡ vừa, 1 bánh cỡ nhỏ là:

\(300000+250000\cdot2+200000=1000000\left(đồng\right)\)

=>bác Lan đủ tiền mua

Bài 2:

a: \(x-0,5=\dfrac{5}{6}\)

=>\(x=\dfrac{5}{6}+\dfrac{1}{2}=\dfrac{5}{6}+\dfrac{3}{6}=\dfrac{8}{6}=\dfrac{4}{3}\)

b: \(\left|x-1\right|+\dfrac{1}{2}=\dfrac{3}{2}\)

=>\(\left|x-1\right|=\dfrac{3}{2}-\dfrac{1}{2}=\dfrac{2}{2}=1\)

=>\(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

\(|x^2|x+\dfrac{3}{4}||=x^2\)

=>\(x^2\cdot\left|x+\dfrac{3}{4}\right|=x^2\)

=>\(\left|x+\dfrac{3}{4}\right|=1\)

=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=1\\x+\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)

|\(x^2\).|\(x+\dfrac{3}{4}\)| |= \(x^2\)

\(x^2\).|\(x+\dfrac{3}{4}\)| = \(x^2\)

\(x^2\).|\(x+\dfrac{3}{4}\)| - \(x^2\) = 0

\(x^2\).(|\(x+\dfrac{3}{4}\)| - 1) = 0

\(\left[{}\begin{matrix}x=0\\\left|x+\dfrac{3}{4}\right|=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x+\dfrac{3}{4}=-1\\x+\dfrac{3}{4}=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\) 

Vậy \(x\) \(\in\) { - \(\dfrac{7}{4}\); 0; \(\dfrac{1}{4}\)}