Bài làm :

Ta có :

 \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{\left(n+1\right)!}\)

\(=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{n}{1.2.3...\left(n+1\right)}\)

\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{n+1-1}{1.2.3...\left(n+1\right)}\)

\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.4..n}-\frac{1}{1.2.3.4...\left(n+1\right)}\)

\(=1-\frac{1}{1.2.3.4...\left(n+1\right)}\)

\(\text{Vì : }\frac{1}{1.2.3.4...\left(n+1\right)}>0\Rightarrow1-\frac{1}{1.2.3.4...\left(n+1\right)}< 1\)

=> Điều phải chứng minh

Ta có : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{\left(n+1\right)!}=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{n}{1.2.3...\left(n+1\right)}\)

\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{n+1-1}{1.2.3....\left(n+1\right)}\)

\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3.4..n}-\frac{1}{1.2.3.4...\left(n+1\right)}\)

\(=1-\frac{1}{1.2.3.4...\left(n+1\right)}< 1\left(\text{đpcm}\right)\)

ai giup vs huhu

a, Áp dụng dãy tỉ số bàng nhau ta có : 

\(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)

\(x=14;y=26\)

b, Áp dụng dãy tỉ số bằng nhau ta có : 

\(\frac{x}{19}=\frac{y}{21}=\frac{2x-y}{2.19-21}=\frac{34}{17}=2\)

\(x=38;y=42\)

áp dụng tính chất dãy tỉ số bằng nhau ta được

\(\frac{x}{7}=\frac{y}{13}=\frac{x+y}{7+13}=\frac{40}{20}=2\)

x=2.7=14

y=2.13=26

vậy x=14    y=26

áp dụng tính chất dãy tỉ số bằng nhau ta được

\(\frac{x}{19}=\frac{y}{21}=\frac{x}{38}=\frac{y}{21}=\frac{x-y}{38-21}=\frac{34}{17}=2\)

x=2.38=76

y=2.21=42

vậy x=76   y=42

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)

Vậy.............

Áp dụng t/c dãy tỉ số bằng nhau 

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Suy ra  \(\left(\frac{a}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)

Ta có ddpcm 

đăng câu hỏi sp mk mk sp lại

              Bài làm :

a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)

b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)

c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)

d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)

e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)

a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)

                                Bài giải

a)\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{3^3\cdot2^3+3^3\cdot2^2+3^3}{-13}=\frac{3^3\left(2^3+2^2+1\right)}{-13}=-3^3=-27\)

b) \(\frac{2^3+3\cdot2^6-4^3}{2^3+3^2}=\frac{8+3\cdot64-64}{8+9}=\frac{8+192-64}{17}=\frac{136}{17}=8\)

c) \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{11}\cdot3^{10}\left(2+2\cdot5\right)}{2^{11}\cdot3^{10}\cdot\left(2\cdot3^2-3\right)}=\frac{12}{18-3}=\frac{12}{15}\)

d) \(\frac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}=\frac{5^5\cdot20^3-5^4\cdot20^3+20^3\cdot20^2\cdot5^2}{5^6\cdot4^5}=\frac{20^3\left(5^5-5^4+5^4\cdot4^2\right)}{20^5\cdot5}\)\(=\frac{5^4\left(5-1+16\right)}{20^2\cdot5}=\frac{5^4\cdot20}{20^2\cdot5}=\frac{5^3}{20}=\frac{5^3}{5\cdot4}=\frac{25}{4}\)

a) \(\frac{3}{4}-\left(\frac{1}{2}:x+\frac{1}{2}\right)=\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{3}{4}-\frac{3}{5}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{15}{20}-\frac{12}{20}\)

\(\Leftrightarrow\frac{1}{2}:x+\frac{1}{2}=\frac{13}{20}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{13}{20}-\frac{10}{20}\)

\(\Leftrightarrow\frac{1}{2}:x=\frac{3}{20}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{3}{20}\)

\(\Leftrightarrow x=\frac{1}{2}.\frac{20}{3}=\frac{10}{3}\)

Vậy: \(x=\frac{10}{3}\)

b) \(3x.\left(\frac{1}{2}.x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\\frac{1}{2}x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\\frac{1}{2}x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=1:\frac{1}{2}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)

Vậy: \(x\in\left\{0;2\right\}\)

c) \(\left(4-x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\2x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{3}{2}\end{cases}}}\)

Vậy: \(x\in\left\{4;\frac{3}{2}\right\}\)

d) \(\frac{4}{-3}=\frac{-12}{x}\)

\(\Leftrightarrow4x=\left(-12\right).\left(-3\right)\)

\(\Leftrightarrow4x=36\)

\(\Leftrightarrow x=9\)

Vậy: \(x=9\)

e) \(\frac{4x}{-3}=\frac{12}{-x}\)

\(\Leftrightarrow4x.\left(-x\right)=12.\left(-3\right)\)

\(\Leftrightarrow-4x^2=-36\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy: \(x\in\left\{3;-3\right\}\)

a,Ta có: 3/4-(1/2:x+1/2)=3/5

                  -(1/2:x+1/2)=3/5-3/4

                  -(1/2:x+1/2)=-3/20

                      1/2:x+1/2=3/20

                             1/2:x=3/20-1/2

                             1/2:x=-7/20

                                   x=1/2:-7/20

                                   x=-10/7

b,Ta có: 3x.(1/2x-1)=0

 Với 3x=0 =>x=0

vói1/2x-1=0

Từ \(\frac{a}{b}=\frac{c}{d}\), áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\)

hay \(\frac{a}{b}=\frac{a+c}{b+d}\)và \(\frac{a}{b}=\frac{a-c}{b-d}\)

                              Trả lời

Từ \(\frac{a}{b}=\frac{c}{d}\), áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}=\frac{a-c}{b-d}\)

hay \(\frac{a}{b}=\frac{a+c}{b+d}\)và \(\frac{a}{b}=\frac{a-c}{b-d}\)

a) x - y = 2 (x + y) x : y

Ta có: 2 (x + y) = x - y

=> 2x + 2y = x - y

=> 2x - x = y - 2y

=> x = -3y

=> x : y = -3

Vì x : y = x - y = 2 (x + y) = -3 nên ta có: x - y = -3 và 2 (x + y) = -3

=> x - y = -3 và x + y = -3/2 (*)

=> x - y + x + y = -3 + 3/2

=> 2x = -9/2

=> x = -9/4

Thay vào (*), ta được:

-9/4 + y = -3/2

=> y = -3/2 + 9/4 = 3/4

Vậy x = -9/4, y = 3/4.

b) x + y = xy = x : y

Ta có: x + y = xy (*)

=> xy - y = x

=> y (x - 1) = x

=> x : y = x - 1

Mà x + y = x : y nên x + y = x - 1

=> x - x + y = -1

=> y = -1

Thay vào (*), ta được: x + (-1) = -1x

x + (-1) = -x

x + x = 1

2x = 1

=> x = 1/2

Vậy x = 1/2, y = -1