Ta có:\(2009^{20}=\left(2009^2\right)^{10}=4036081^{10}< 20092009^{10}\)

Vậy \(2009^{20}< 20092009^{10}\)

chẳng nhìn thấy j cả! Thông cảm mk bị cận!

Theo mình nghĩ thì đề thiếu là tam giác ABC vuông tại A nhé!

Bạn xem lại đề!:)

số hay hình bn, trang mấy. Bn phải ghi rõ ra chứ

kẻ đường thẳng OK sao cho OK // a

Ta có góc A+KOA=180^o( hai góc trong cùng phía bù nhau)

=> góc KOA=180^o-110^o=70^o

=> góc KOB=140^o - 70^o = 70^o

Mà KOB+B=70^o+110^o=180^o

=> OK//b Mà OK//a; OK//b

=>a//b. tick giùm tui đi, please V_V

Giải:

a) Có: \(0,\left(37\right)=0,373737373737...\)

\(0,\left(62\right)=0,626262626262...\)

\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999999...\)

Mà \(0,9999999999999...\simeq1\)

Hay \(0,\left(9\right)=1\)

Vậy \(0,\left(37\right)+0,\left(62\right)=1\).

b) \(0,\left(33\right).3=0,99999...=0,\left(9\right)=1\)

Vậy \(0,\left(33\right).3=1\).

Chúc bạn học tốt!!!

\(a)0,\left(37\right)=0,37373737....\)

\(0,\left(62\right)=0,62626262....\)\(\Leftrightarrow0,\left(37\right)+0,\left(62\right)=0,99999999....\)

Mà \(0,99999999....\simeq1\)

hoặc \(0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(37\right)+\left(0,62\right)=1\)

\(b)0,\left(33\right).3=1\)

\(\Leftrightarrow0,99999999....=0,\left(9\right)\simeq1\)

\(\Rightarrow0,\left(33\right).3=1\)

Chúc bạn học tốt!

\(\left(\dfrac{-5}{13}\right)^{2017}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(-\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(-\dfrac{5}{13}\right)\cdot\left[\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}\right]=\left(-\dfrac{5}{13}\right)\cdot1^{2016}=\left(-\dfrac{5}{13}\right)\cdot1=-\dfrac{5}{13}\)

k giống đâu bạn ạ

\(xy-x-y+1=0\)

\(\Rightarrow x.\left(y-1\right)-\left(y-1\right)=0\)

\(\Rightarrow\left(y-1\right).\left(x-1\right)=0\)

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

Vậy \(x=y=1\)

Chúc bạn học tốt!!!

Tìm x,y biết:

xy-x-y+1=0

=> x(y-1)-y=0-1

=> x(y-1)- (y-1)= (-1)

=> (y-1)(x-1)=(-1)

\(\Rightarrow\left[{}\begin{matrix}y-1=1;x-1=-1\\y-1=-1;x-1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=2;x=0\\y=0;x=2\end{matrix}\right.\)

Ta có: \(\left|x-1\right|+\left|x-5\right|=\left|x-1\right|+\left|5-x\right|\)

Nhận thấy: \(\left[{}\begin{matrix}\left|x-1\right|\ge x-1\\\left|5-x\right|\ge5-x\end{matrix}\right.\)

\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge x-1+5-x\)

\(\Rightarrow\left|x-1\right|+\left|5-x\right|\ge4\)

Dấu \("="\) xảy ra khi:

\(\left[{}\begin{matrix}x-1\ge0\\5-x\ge0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge1\\x\le5\end{matrix}\right.\) \(\Rightarrow1\le x\le5\)

Vậy \(1\le x\le5.\)

Cho mk thêm cái ạ:

\(x\in\left\{1;2;3;4;5\right\}\)

Vậy \(x\in\left\{1;2;3;4;5\right\}\)