`Answer:`

Đặt \(N=\frac{9}{10}+\frac{9}{10^2}+\frac{9}{10^3}+...+\frac{9}{10^{10}}\)

\(\Rightarrow\frac{N}{9}=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{10}}\)

Ta có \(\frac{N}{90}=\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+...+\frac{1}{10^{11}}\)

\(\Rightarrow\frac{N}{9}-\frac{N}{90}=\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{10}}\right)-\left(\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+...+\frac{1}{10^{11}}\right)\)

\(\Rightarrow\frac{10N}{90}-\frac{N}{90}=\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^{10}}-\frac{1}{10^2}-\frac{1}{10^3}-\frac{1}{10^4}-...-\frac{1}{10^{11}}\)

\(\Rightarrow\frac{9N}{90}=\frac{1}{10}-\frac{1}{10^{11}}\)

\(\Rightarrow\frac{N}{10}=\frac{10^{10}-1}{10^{11}}\)

\(\Rightarrow N=\frac{10^{10}-1}{10^{10}}\)

Do ƯCLN ( a , b ) = 15 => a = 15 . m ; b = 15 . n  ( m,n) = 1

=> BCNN ( a, b ) = 15 . m . n = 300

=> m . n = 300 : 15 = 20

Nếu a > b thì m > n do ( m;n ) = 1 =>  m = 20 ; n = 1 hoặc m = 5; n = 4

+Với m = 20 , n = 1 thì a =15 . 20 = 300 ; b = 15 . 1 = 15

+Với m = 5 , n = 4 thì a = 15 . 5 = 75 ; b = 15 . 4 = 60

Vậy các cặp giá trị (m;n) thỏa mãn đề bài là : ( 300 ; 15 ) ; ( 15 ; 300 ) ; ( 75 ; 60 ) ; ( 60 ; 75 )

em lên câu hỏi tương tự có đấy

a) Ngày đầu Nam đọc được là:

480 . 5/8 = 300 ( trang )

Ngày thứ hai Nam đọc được là:

(480 - 300 ) . 60% = 72 ( trang )

b) Tỉ số phần trămcủa số trang sách còn lại sau hai ngày so với số trang của cuốn sách là:

[( 480 - 300 - 72 ) : 480] : 100 % = 22,5%

                                                     Đáp số : a) Ngày đầu : 300 trang 

                                                                    Ngày thứ hai : 72 trang 

                                                                    b) 22,5%\

cho mk nekkk

- sự bay hơi là sự chuyển từ thể lỏng sang thể hơi

- ví dụ: nước được đun nóng hay làm lạnh thì ta có thể thấy sự bay hơi

- tốc độ bay hơi của một chất lỏng phụ thuộc vào nhiệt độ, gió, và mặt thoáng cảu chất lỏng

+ thí nghiệm chứng tỏ tốc độ bay hơi của nước phụ thuộc vào gió:

một chiếc đĩa chứa lượng nước như nhau. một cái để ở nơi có nhiều gió. một cái để trong phòn kín 

=> sau một thời gian thì đĩa ở nơi có gió tất nhiên sẽ bay hơi nhiều hơn

+ thí nghiệm chứng tỏ tốc độ bay hơi của nước phụ thuộc vào nhiệt độ:

một chiếc đĩa chứa lượng nước như nhau đều để trong phong kín. một cái để ko. một cái đun trên đèn cồn

=>sau một thời gian thì đĩa đun nước bốc hơi nhanh hơn nên cạn dần, ít nước hơn đĩa kia

+ thí nghiệm chứng tỏ tốc độ bay hơi của nước phụ thuộc vào mặt thoáng của chất lỏng:

một chiếc đĩa chứa lượng nước như nhau đều để trong phong kín. nhưng một chiếc thì to và rộng. một chiếc nhỏ và chật hẹp

=>sau một thời gian thì đĩa có mặt thoán rỗng rã thì tất nhiên sẽ bay hơi nhiều hơn 

Chúc bạn học tốt >.<

ta có (x-1)(y+2)=5

do x,y nguyên nên x-1, y+2 nguyên

=> \(x-1,y+2\inƯ\left(5\right)\). Nên ta có bảng sau

vậy ......

bạn tự kết luận nhé

tk mk nha 

*****Chúc bạn học giỏi*****

Ta có : 5 = 1 x 5 = (-1) x (-5) = 5 x 1 = (-5) x (-1)

Với : \(\hept{\begin{cases}x-1=1\\y+2=5\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}\)

\(\hept{\begin{cases}x-1=-1\\y+2=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=-7\end{cases}}\)

\(\hept{\begin{cases}x-1=5\\y+2=1\end{cases}}\Rightarrow\hept{\begin{cases}x=6\\y=-1\end{cases}}\)

\(\hept{\begin{cases}x-1=-5\\y+2=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-4\\y=-3\end{cases}}\)

Vậy x = 2, y = 3 ; x = 0, y = -7; x = 6, y = -1; x = -4, y = -3

=>x^2-6x+9=\(\frac{4\left(4^{16}+16\right)}{4^{16}+16}\)
=>x^2-6x+9=4

=>x^2-6x+5=0

=>(x^2-x)-(5x-5)=0

=>x(x-1)-5(x-1)=0

=>(x-1)(x-5)=0

=>\(\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

vậy x=1;x=5

a) \(3,5x-25=60.\frac{3}{4}\Leftrightarrow3,5x-25=45\)

\(\Leftrightarrow3,5x=45+25\Leftrightarrow3,5x=70\Leftrightarrow x=\frac{70}{3,5}=20\)

b) \(-1\frac{1}{2}x+\frac{1}{5}=2\frac{3}{10}\Leftrightarrow\frac{-3}{2}x=\frac{23}{10}\Leftrightarrow x=\frac{23}{10}:\frac{-3}{2}=\frac{-23}{15}\)

a)\(3,5x-25=60:\frac{3}{4}=80\)

\(3,5x=80+25\)

\(3,5x=105\)

\(x=105:3,5\)

\(x=30\)

b) \(\frac{-3}{2}x+\frac{1}{5}=\frac{23}{10}\)

\(\frac{-3}{2}x=\frac{23}{10}-\frac{1}{5}=\frac{21}{10}\)

\(x=\frac{21}{10}:\frac{-3}{2}\)

\(x=\frac{-7}{5}\)

21342+4214= 25556

lâu rồi ko ns chuyện nhỉ linh

\(P=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}+\frac{15}{4.43}+\frac{13}{43.8}\)

\(\Leftrightarrow\)\(\frac{1}{7}P=\frac{1}{7}\left(\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}+\frac{15}{4.43}+\frac{13}{43.8}\right)\)

                   \(=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}+\frac{15}{28.43}+\frac{13}{43.56}\)

                   \(=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}+\frac{1}{28}-\frac{1}{43}+\frac{1}{43}-\frac{1}{56}\)

                    \(=\frac{1}{2}-\frac{1}{56}=\frac{27}{56}\)

\(\Leftrightarrow\)\(P=\frac{27}{56}:\frac{1}{7}=3\frac{3}{8}\)\(>3\)  (ĐPCM)

1/7p= 5/2.7+4/7.11+...+13/43.56

1/7p=1/2-1/7+1/7-1/11+...+1/43-1/56

1/7p=1/2-1/56

1/7p=27/56

suy ra p=27/56.7

p=189/56>3 suy ra đéo phải chứng minh

ai mà tính được ??

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