a) \(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(A=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)

\(A=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(A=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(A=\dfrac{1-3}{1+5}\)

\(A=-\dfrac{2}{6}\)

\(A=-\dfrac{1}{3}\)

b) \(B=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(B=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\cdot\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)

\(B=\dfrac{1}{4}+\dfrac{3\cdot\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\cdot\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(B=\dfrac{1}{4}+\dfrac{3}{4}\)

\(B=\dfrac{1+3}{4}\)

\(B=\dfrac{4}{4}\)

\(B=1\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

a) => (2x - 15)⁵ - (2x - 15)³ = 0

(2x-15)³ [(2x-15)²-1]=0

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

b) x + 2x + 3x + ... + 2022x = 2022.2023

x(1 + 2 + 3 +...+ 2022)= 2022.2023 (1)

Đặt A = 1+2+3+...+2022

Số số hạng trong A là: (2022 - 1): 1 + 1 = 2022 (số)

Tổng A bằng: \(\dfrac{\left(2022+1\right).2022}{2}=\dfrac{2022.2023}{2}\)

Thay A vào (1) , ta  được:

x.\(\dfrac{2022.2023}{2}=2022.2023\)

=> x = 2

\(x+2x+3x+...+2022x=2022\cdot2023\)

\(\Rightarrow x\cdot\left(1+2+3+...+2022\right)=2022\cdot2023\)  

\(\Rightarrow x\cdot\left(2022+1\right)\cdot\left[\left(2022-1\right):1+1\right]:2=2022\cdot2023\)

\(\Rightarrow x\cdot2023\cdot\left(2022-1+1\right):2=2022\cdot2023\)

\(\Rightarrow x\cdot2023\cdot2022:2=2022\cdot2023\)

\(\Rightarrow x:2=2022\cdot2023:2022\cdot2023\)

\(\Rightarrow x:2=1\)

\(\Rightarrow x=1\cdot2\)

\(\Rightarrow x=2\)

a) Ta có : 34 + ( 21  - x ) = ( 3747 - 30 ) - 3746

<=> 34 + 21 - x                =  3717 - 3746

<=> 55- x                           = -29

<=>     -x                              = -29 - 55

<=>  - x                                   = -84

<=> x                                    = 84

Vậy x = 84

b)Ta có  | x + 2 | + 21 = 25

<=> | x + 2 |                = 25 - 21

<=> | x + 2 |                = 4

<=> \(\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

Vậy x = { 2 ; -6 }

Mn giúp mik với nhé. Mình đag cần gấp.

Còn ai onl ko giúp mình với nhé 

Mn ai còn onl thì giúp mình với nhé.

\(BC=AC-AB=60\left(km\right)\\ \Rightarrow\%_{\dfrac{s_{BC}}{s_{AB}}}=\dfrac{60}{80}\times100\%=75\%\)

3:

a: Xét ΔABD và ΔAED có

AB=AE

góc BAD=góc EAD

AD chung

=>ΔABD=ΔAED

=>DB=ED

b; Xét ΔDBK và ΔDEC có

góc DBK=góc DEC

DB=DE

góc BDK=góc EDC

=>ΔDBK=ΔDEC

c: AB+BK=AK

AE+EC=AC

mà AB=AE; BK=EC

nên AK=AC

=>ΔAKC cân tại A

d: ΔAKC cân tại A

mà AD là phân giác

nên AD vuông góc KC

Bài 9:

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{z}{-17}=\dfrac{t}{9}=-2\)

=>x=-10; y=6; z=34; t=-18

Bài 10:

\(\Leftrightarrow\dfrac{8}{x}=\dfrac{y}{21}=\dfrac{40}{z}=\dfrac{16}{t}=\dfrac{u}{111}=\dfrac{4}{3}\)

=>x=6; y=28; z=30; t=12; u=148