có dây  vào nick tớ mà chơi

Gửi tên nick đi mkm sẽ kbn

\(\frac{x}{7}=\frac{x+36}{35}\)

\(\Rightarrow x.35=7.\left(x+36\right)\)

\(\Rightarrow35x=7x+252\)

\(\Rightarrow35x-7x=252\)

\(\Rightarrow28x=252\)

\(\Rightarrow x=9\)

/ laf chia af

Ta có: \(\frac{-3}{4}=\frac{-6}{x}\Rightarrow x=\frac{4.\left(-6\right)}{-3}=8\)

 Thế x = 8 \(\Rightarrow\frac{-6}{8}=\frac{y}{20}\Rightarrow y=\frac{\left(-6\right).20}{8}=-15\)

Thế y = -15 \(\Rightarrow\frac{-15}{20}=\frac{-15}{z}\Rightarrow z=\frac{20.\left(-15\right)}{-15}=20\)

Vậy x = 8 ; y = -15 ; z = 20

a) (-4)² + 21 + 2x = 3⁶ : 3³

16 + 21 + 2x = 3³

37 + 2x = 27

2x = 27 - 37

2x = -10

x = -10 : 2

x = -5

b) 8 - 2x = -12

2x = 8 - (-12)

2x = 20

x = 20 : 2

x = 10

c) (79 + x) - 43 = -(17 - 55)

79 + x - 43 = 38

36 + x = 38

x = 38 - 36

x = 2

mik c.ơn 😫😫😫😘😘😘😩🤗😚😌😌

a) 1+(-3)+5+(-7)+9+(-11)+13+(-15)

= [1+(-3)]+[5+(-7)]+[9+(-11)]+[13+(-15)]

=(-2)+(-2)+(-2)+(-2)

=(-2).4

=-8

b) (-1)+3+(-5)+7+...+(-2005)+2007

=[(-1)+3]+[(-5)+7]+...+[(-2005)+2007]

=2+2+...+2

Từ 1 đến 2007 có : (2007-1)/2+1=1004 số

Mà cứ 2 số ta được 1 cặp

=>1004/2=502 cặp

=>2.502=1004

Câu c) làm tương tự câu b) nha

Nhớ **** cho mình nha ^-^

cô ơi cho con hỏi 

11 mũ 5 chia cho 11 mũ n trừ 4 bằng 11 mũ 5 

giúp tính bài giải cho con nhé

a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)

\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)

c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)

\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)

\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)

d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)

\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)

\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)

\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)

e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)

\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)

g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

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

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé