Ta có : $2^x+2^{x+1}+2^{x+2}+...+2^{x+2015}=2^{2019}-8$

$\iff 2^x\left(1+2+2^2+...+2^{2015}\right)=2^{2019}-8$ (1)

Đặt : $A=1+2+2^2+...+2^{2015}$

$\Rightarrow 2A=2+2^2+2^3+...+2^{2016}$

$\Rightarrow 2A-A=\left(2+2^2+2^3+...+2^{2016}\right)-\left(1+2+2^2+...+2^{2015}\right)$

$\Rightarrow A=2^{2016}-1$

Khi đó (1) trở thành :

$2^x\left(2^{2016}-1\right)=2^{2019}-2^3$

$\iff 2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)$

$\iff 2^x=2^3\left(2^{2016}-1\ne 0\right)$

$\iff x=3$

Vậy : $x=3$

$2^x+2^{x+1}+...+2^{x+2015}=2^{2019}-8$

$\to 2^x.1+2^x.2+....+2^x.2^{2015}=2^{2019}-8$

$\to 2^x.(1+2+...+2^{2015})=2^{2019}-8$

Đặt:

$A=1+2+...+2^{2015}$

$2A=2.(1+2+...+2^{2015})$

$2A=2+2^2+...+2^{2016}$

$2A-A=(2+2^2+...+2^{2016})-(1+2+...+2^{2015})$

$A=2+2^2+...+2^{2016}-1-2-...-2^{2015}$

$A=2^{2016}-1$

Nên:

$2^x.(1+2+...+2^{2015})=2^{2019}-8$

$\to 2^x.(2^{2016}-1)=2^{2019}-8$

$\to 2^x=(2^{2019}-8):(2^{2016}-1)$

$\to 2^x=\frac{2^{2019}-8}{2^{2016}-1}$

$\to 2^x=\frac{2^3.(2^{2016}-1)}{2^{2016}-1}$

$\to 2^x=2^3$

$\to x=3$

Vậy $x=3.$

Bài 5:

\(\dfrac{3}{4}=\dfrac{x}{20}=\dfrac{21}{y}\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{20}=\dfrac{3}{4}\\\dfrac{21}{y}=\dfrac{3}{4}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=20\cdot\dfrac{3}{4}=15\\y=21\cdot\dfrac{4}{3}=7\cdot4=28\end{matrix}\right.\)

Bài 4:

a: \(\dfrac{3}{5}=\dfrac{3\cdot9}{5\cdot9}=\dfrac{27}{45}\)

b: Đề sai rồi bạn

Bài 3:

a: \(21\cdot21=441\)

\(49\cdot9=441\)

=>\(21\cdot21=49\cdot9\)

=>\(\dfrac{21}{9}=\dfrac{49}{21}\)

b: \(\dfrac{-24}{34}=\dfrac{-24:2}{34:2}=\dfrac{-12}{17}\)

\(\dfrac{-60}{85}=\dfrac{-60:5}{85:5}=\dfrac{-12}{17}\)

Do đó: \(\dfrac{-24}{34}=\dfrac{-60}{85}\)

Bài 2:

\(15cm=\dfrac{3}{20}m\)

\(40mm=\left(\dfrac{40}{1000}\right)m=\dfrac{1}{25}m\)

Bài 1:

a: \(\left(-17\right):8=\dfrac{-17}{8}\)

b: \(\left(-8\right):\left(-9\right)=\dfrac{-8}{-9}\)

a)

\(\left(2-x\right)^3=\left(2-x\right)^5\\ \Rightarrow\left(2-x\right)^3-\left(2-x\right)^5=0\\ \Rightarrow\left(2-x\right)^3\left[1-\left(2-x\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(2-x\right)^3=0\\1-\left(2-x\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2-x=0\\\left(2-x\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\2-x=\pm1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=3\end{matrix}\right.\)

\(2xy-x-y=2\)

\(\Rightarrow x\left(2y-1\right)-y=2\)

\(\Rightarrow x\left(2y-1\right)=y+2\)

\(\Rightarrow2x\left(2y-1\right)=2y+4\) 

\(\Rightarrow2x\left(2y-1\right)-\left(2y-1\right)=5\)

\(\Rightarrow\left(2x-1\right)\left(2y-1\right)=5=1\cdot5=5\cdot1=-1\cdot-5=-5\cdot-1\) 

Ta có bảng sau: 

Bài `1`

`a,x/15+7/20=73/60`

`=> x/15= 73/60 - 7/20`

`=> x/15=13/15`

`=> x=13`

`b,9/x + (-6)/35 =33/70`

`=> 9/x = 33/70 - (-6)/35`

`=>9/x = 33/70 + 6/35`

`=> 9/x=9/14`

`=>x=14`

Bài `2`

`a,` \(\dfrac{12}{13}=\dfrac{12\cdot101}{13\cdot101}=\dfrac{1212}{1313}=\dfrac{12\cdot\left(-10101\right)}{13\cdot\left(-10101\right)}=\dfrac{-121212}{-131313}=\dfrac{121213:10101}{131313:10101}=\dfrac{12}{13}\)

`b,` 

\(-\dfrac{22}{121}=\dfrac{-2\cdot2}{11\cdot2}=-\dfrac{4}{-22}\\ -\dfrac{10}{55}=\dfrac{-10:5}{55:5}=-\dfrac{2}{11}=-\dfrac{22}{121}\)

⇒2x-6y=6

ta có: 6=1.6=2.3=-2.-3=-1.-6

xong xét từng trường hợp ra nhé.

Chia hết cho 5 mà không chia hết cho 2 => b=5

Ta có, tổng các chữ số của số trên là: 8 + 2 + 5 + a + 2 + b = 8 + 2 + 5 + a + 2 + 5 = 22 + a 

Để số trên chia hết cho 9 thì (22+a) chia hết cho 9, vậy a = 5 (Để: 22+a=27, và 27:9=3)

Vậy: Với a=5;b=5 thì số 825a2b chia hết cho cả 5 và 9 nhưng không chia hết cho 2