\(\frac{3}{4}+\frac{3}{28}+\frac{3}{70}+\frac{3}{130}+....+\frac{3}{418}+\frac{3}{550}\)

\(\Leftrightarrow\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{19.22}+\frac{3}{22.25}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{22}+\frac{1}{22}-\frac{1}{25}\)

\(\Leftrightarrow\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)

Nhớ k cho m nhé!

\(3\times\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)

=> \(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}=\frac{20}{13}\)

=> \(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}=\frac{20}{13}\)

=> \(\frac{x}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\left(1-\frac{1}{13}\right)=\frac{20}{13}\)

=> \(\frac{x}{3}\cdot\frac{12}{13}=\frac{20}{13}\)

=> \(\frac{x}{3}=\frac{20}{13}:\frac{12}{13}=\frac{20}{13}\cdot\frac{13}{12}=\frac{5}{3}\)

=> x = 5

\(3\cdot\left(\frac{x}{4}+\frac{x}{28}+\frac{x}{70}+\frac{x}{130}\right)=\frac{60}{13}\)

\(3\cdot\left(\frac{x}{1\cdot4}+\frac{x}{4\cdot7}+\frac{x}{7\cdot10}+\frac{x}{10\cdot13}\right)=\frac{60}{13}\)

\(3\left(x-3\right)\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\left(1-\frac{1}{13}\right)=\frac{60}{13}\)

\(\left(3x-9\right)\cdot\frac{12}{13}=\frac{60}{13}\)

\(3x-9=\frac{\frac{60}{13}}{\frac{12}{13}}\)

\(3x-9=5\)

\(3x=5+9\)

\(3x=14\)

\(x=\frac{14}{3}\approx4,667\)

b) D = \(\frac{3}{4}+\frac{3}{8}+\frac{3}{70}+\frac{3}{130}+\frac{3}{208}+\frac{3}{304}\)

    D =  \(3\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)

    D = \(3\left(\frac{1}{1x4}+\frac{1}{4x7}+\frac{1}{7x10}+\frac{1}{10x13}+\frac{1}{13x16}+\frac{1}{16x19}\right)\)

    D = \(\frac{1}{1}-\frac{1}{19}=\frac{18}{19}\)

Chắc vậy

CÁCH LÀM NHƯ SAU : 

(7/28 + 1/28) + 1/70 + 1/130 + 1/x.(x+3)

8/28 + 1/70 +1/130 +1/x.(x+3)

2/7+1/70+1/130+1/x.(x+3)

(20/70 +1/70)+1/130+1/x.(x+3)

3/10+1/130+1/x.(x+3)

39/130+1/130+1/x.(x+3)

4/13+1/x.(x+3)

Đến đây bn tự làm hộ mình vớ. chúc hok tốt k cho mình nhé

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{x\left(x+3\right)}\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}\left(\frac{12}{13}+\frac{1}{x}-\frac{1}{x+3}\right)\)

\(=\frac{1}{3}.\frac{12}{13}+\frac{1}{3}.\frac{1}{x}-\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}-\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3x+3}\)

\(=\frac{4}{13}+\frac{1}{3x}=\frac{1}{3}.\frac{1}{x+3}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3x}\)

\(=\frac{4}{13}=\frac{1}{3}.\frac{1}{x+3}-\frac{1}{3}.\frac{1}{x}\)

\(=\frac{4}{13}=\frac{1}{3}\left(\frac{1}{x+3}-\frac{1}{x}\right)\)

\(=\frac{4}{13}:\frac{1}{3}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{1}{x+1}-\frac{1}{x}\)

\(=\frac{12}{13}=\frac{x-\left(x+1\right)}{\left(x+1\right)x}\)

\(=\frac{12}{13}=-\frac{1}{x^2+x}\)

\(\Leftrightarrow=12\left(x^2+x\right)=13.\left(-1\right)\)

\(=12\left(x^2+x\right)=-13\)

\(=x^2+x=-\frac{13}{12}\)

\(=x\left(x+1\right)=-\frac{13}{12}\)

.... Chiụ 

b, (y+1) + (y+2) + (y+3) + .... + (y+99) = 6138

(y+y+y+y+....+y) + (1+2+3+...+99) = 6138

99.y + 4950 = 6138

99.y = 6138 - 4950

99.y = 1188

y = 1188 : 99

x = 12

a) \(\left(y+1\right)+\left(y+4\right)+\left(y+7\right)+....+\left(y+28\right)=155155\)

\(\Rightarrow\left(y+y+...+y\right)+\left(1+4+7+...+28\right)=155155\)

\(\Rightarrow10x+145=155155\)

\(\Rightarrow10x=155010\)

\(\Rightarrow x=15501\)

Vậy x = 15501

b) \(\left(y+1\right)+\left(y+2\right)+..+\left(y+99\right)=6138\)

\(\Rightarrow\left(y+y+...+y\right)+\left(1+2+3+...+99\right)=6138\)

\(\Rightarrow99x+4950=6138\)

\(\Rightarrow99x=1188\)

\(\Rightarrow x=12\)

Vậy x = 12

562m = 0,562km

27,65m = 27m 65cm

Bài 4:

10/7 : y = 1/4 + 3/5

10/7 : y = 17/20

y = 10/7 : 17/20

y = 200/119

HT~

562m = 0,562km

27,65m = 27m 650cm

Bài 4 :

\(10/7 : y = 1/4 + 3/5\)

 \(10/7 : y = 17/20\)

              \(y = 10/7 : 17/20\)

              \(y = 200/119\)

\(\frac{3}{34}+\frac{34}{34}.\frac{3}{4}=\frac{3}{34}+1.\frac{3}{4}=\frac{3}{34}+\frac{3}{4}=\frac{57}{68}\)

\(\frac{23}{3}.\frac{56}{6}+\frac{86}{78}=\frac{23}{3}.\frac{28}{3}+\frac{43}{39}=\frac{644}{9}+\frac{28}{3}=\frac{728}{9}\)

\(\frac{3}{45}:\frac{1}{4}=\frac{1}{15}.4=\frac{4}{15}\)

\(\frac{5}{34}-\frac{3}{6}=\frac{5}{34}-\frac{1}{2}=\frac{3}{4}\)