a: Xét ΔDAC và ΔDMB có

DA=DM

\(\widehat{ADC}=\widehat{MDB}\)(hai góc đối đỉnh)

DC=DB

Do đó: ΔDAC=ΔDMB

=>\(\widehat{DCA}=\widehat{DBM}\)

=>CA//BM

b: Xét ΔDNC và ΔDKB có

\(\widehat{DCN}=\widehat{DBK}\)

DC=DB

\(\widehat{NDC}=\widehat{KDB}\)(hai góc đối đỉnh)

Do đó: ΔDNC=ΔDKB

=>DN=DK

=>D là trung điểm của NK

\(9^6:3^2=9^6:9=9^5\)

9^6 : 3^2 = 9^6 : 9 = 9^5

a: Thời gian dự định sẽ đi hết quãng đường là:

120:50=2,4(giờ)=2h24p

Nếu đúng dự định thì ô tô sẽ đến B lúc:

7h+2h24p=9h24p

b: Đặt AC=x

BC=AB-AC=120-x(km)

Thời gian ô tô đi hết quãng đường AC là \(\dfrac{x}{50}\left(giờ\right)\)

Thời gian ô tô đi hết quãng đường BC là: \(\dfrac{120-x}{60}\left(giờ\right)\)

Ô tô đến B sớm hơn dự kiến 5p nên ta có: \(\dfrac{x}{50}+\dfrac{1}{12}+\dfrac{120-x}{60}=2,4-\dfrac{1}{12}\)

=>\(\dfrac{x}{50}+\dfrac{120-x}{60}=2,4-\dfrac{1}{6}=\dfrac{12}{5}-\dfrac{1}{6}=\dfrac{72-5}{30}=\dfrac{67}{30}\)

=>\(\dfrac{6x+5\left(120-x\right)}{300}=\dfrac{670}{300}\)

=>6x+5(120-x)=670

=>x+600=670

=>x=70(nhận)

Vậy: Độ dài quãng đường AC là 70km

Sửa đề: \(\dfrac{5^2}{1\cdot6}+\dfrac{5^2}{6\cdot11}+...+\dfrac{5^2}{96\cdot101}\)

\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{96\cdot101}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\)

\(=5\left(1-\dfrac{1}{101}\right)=5\cdot\dfrac{100}{101}=\dfrac{500}{101}\)

\(\dfrac{5^2}{6.1}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right) \\ =5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ =5.\left(1-\dfrac{1}{101}\right)\\ =5.\dfrac{100}{101}\\ =\dfrac{500}{101}\)

a: \(4^n=256\)

=>\(4^n=4^4\)

=>n=4

b: \(9^{5n-8}=81\)

=>5n-8=2

=>5n=10

=>n=2

c: \(3^{n+2}:27=3\)

=>\(3^{n+2}=27\cdot3=81=3^4\)

=>n+2=4

=>n=2

d: 

\(8^{n+2}\cdot2^3=8^5\)

=>\(8^{n+2}=8^5:8=8^4\)

=>n+2=4

=>n=2

`4^n = 256`

`=> 4^n = 4^4`

`=> n = 4`

Vậy `n = 4`

\(9^{5n-8}\) `= 81`

=> \(9^{5n-8}\) `= 9^2`

=> `5n-8=2`

=> `5n=10`

=> `n=2`

Vậy `n=2`

\(3^{n+2}:27=3\)

=> \(3^{n+2}=81\)

=> \(3^{n+2}=3^4\)

`=> n + 2 = 4`

`=> n= 2`

Vậy `n = 2`

\(8^{n+2}.2^3=8^5\)

=> \(8^{n+2}.8=8^5\)

=> \(8^{n+2}=8^5:8\)

=> \(8^{n+2}=8^4\)

=> `n+ 2 = 4`

=> `n = 2`

Vậy `n=2`

`(x - 1954) . 5 = 50`

`=> x - 1954 = 50 : 5`

`=> x - 1954 = 10`

`=> x = 10 + 1954`

`=> x = 1964`

Vậy `x = 1964`

`x - 152 : 2 = 46`

`=> x - 76 = 46`

`=> x = 46 + 76`

`=> x = 122`

Vậy `x = 122`

`70 - 5 (x - 3)=45`

`=> 5(x-3)=70-45`

`=> 5(x-3)=25`

`=> x-3=25 : 5`

`=> x - 3 = 5`

`=> x = 5+3`

`=> x = 8`

Vậy `x = 8`

`60-3(x-2)=51`

`=> 3(x - 2)=60-51`

`=> 3(x-2) = 9`

`=> x - 2 = 9:3`

`=> x -2 = 3`

`=> x = 3+2`

`=> x = 5`

Vậy `x = 5`

`10+2x=45.43`

`=> 10 +2x=1935`

`=> 2x = 1935 - 10`

`=> 2x=1925`

`=> x = 1925 : 2`

`=> x =` \(\dfrac{1925}{2}\)

Vậy `x =` \(\dfrac{1925}{2}\)

`4x - 20=25:23`

`=> 4x - 20 =` \(\dfrac{25}{23}\)

`=> 4x =`  \(\dfrac{25}{23}\) `+ 20`

`=> 4x =` \(\dfrac{485}{23}\)

`=> x =` \(\dfrac{485}{23}:4\)

`=> x =` \(\dfrac{485}{23}.\dfrac{1}{4}\)

`=>x=` \(\dfrac{485}{92}\)

Vậy `x =` \(\dfrac{485}{92}\)

c) ( x - 1954 ).5 = 50

 x - 1954 = 10

x = 1964

Vậy...

d) x - 152:2 = 46

x - 152 = 92

x = 244

Vậy...

e) 70 - 5 ( x- 3) = 45

65 ( x - 3 ) = 45

x - 3 = 9/13

x = 48/13

Vậy...

g) 60 - 3 ( x - 2 ) = 51

57 ( x - 2 ) = 51

x - 2 =17/19

x = 55/19

Vậy...

f) 10 + 2x = 45.43

10 + 2x = 1935

2x = 1925

x = 1925/2

Vậy...

h) 4x - 20 = 25:23

4x - 20 = 25/23

4x = 485/23

x = 485/92

Vậy...

Ta có:

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Áp dụng:

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{625}\right)\)

\(=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{25^2}\right)\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{24.26}{25^2}\)

\(=\dfrac{1.2.3...24}{2.3.4...25}.\dfrac{3.4.5...26}{2.3.4...25}=\dfrac{1}{25}.\dfrac{26}{2}=\dfrac{13}{25}\)

\(=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{575}{576}.\dfrac{624}{625}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{24.26}{25.25}\)

\(=\dfrac{\left(1.2...24\right).\left(3.4...26\right)}{\left(2.3...25\right).\left(2.3...25\right)}\)

\(=\dfrac{1.26}{25.2}=\dfrac{1.2.13}{25.2}=\dfrac{13}{25}\)

\(\left(x-3\right)^2=\left(1-3x\right)^2\)

=>\(\left(3x-1\right)^2-\left(x-3\right)^2=0\)

=>\(\left(3x-1-x+3\right)\left(3x-1+x-3\right)=0\)

=>(2x+2)(4x-4)=0

=>\(\left[{}\begin{matrix}2x+2=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Cách 1: Chỉ ra tính chất đặc trưng 

\(B=\left\{x=2k;k\in N|k\le6\right\}\)

Cách 2: liệt kê

\(B=\left\{0;2;4;6;8;10;12\right\}\)

Cách 1: Liệt kê: 

`B =` {`0;2;4;6;8;10;12`}

Cách 2: Đặc trưng:

`B =` {`x` thuộc `N | x ⋮ 2`` và `x ≤13`}