{50:5-45:5}x7

={10-9}x7

=1x7

=7

d,

=6^2x10:{780:[1000-15,625+35x14]}

=6^2x10:{780:[1000-15,625+490]}

=6^2x10:{780:1474,375}

=6^2x10:0,52903772785

=36x10:0,52903772785

=360:0,52903772785

=608,480769232

khum bt câu d, làm đúng chưa?

chắc sai á

\(a,\left\{\left[\left(37+13\right):5\right]-45:5\right\}.7\)

\(=\left\{\left[50:5\right]-45:5\right\}.7\)

\(=\left\{10-45:5\right\}.7\)

\(=\left\{10-9\right\}.7\)

\(=1.7=7\)

\(6^2.10:\left\{780:\left[10^3-\left(2.5^3+35.14\right)\right]\right\}\)

\(=36.10:\left\{780:\left[1000-\left(2.125+35.14\right)\right]\right\}\)

\(=360:\left\{780:\left[1000-\left(250+490\right)\right]\right\}\)

\(=360:\left\{780:\left[1000-740\right]\right\}\)

\(=360:\left\{780:260\right\}\)

\(=360:3=120\)

số đó là

228

ai k mình

mình k lại cho

35+62+14+52+65= ( 35 + 65 ) + 62 + 14 + 52 = 228

35+62+14+52+65=228

\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)

\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)

\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)

\(M=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+...+1-\frac{1}{9999}\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\right)\)

\(M=\left(1+1+1+...+1\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)(Có (99 - 1): 2+ 1 = 50 số 1)

\(M=50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\right)\)

\(M=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(M=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{5050-100}{101}=\frac{4950}{101}\)

1 

2

Đâu rồi

\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(=\frac{3-1}{3}+\frac{15-1}{15}+\frac{35-1}{35}+\frac{63-1}{63}+\frac{99-1}{99}\)

\(=1-\frac{1}{3}+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}\)

\(=5+\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(=5+\frac{1}{2}\left(1-\frac{1}{11}\right)\)

\(=5+\frac{5}{11}=\frac{60}{11}\)

Sửa đề: \(98+99+\dfrac{142}{144}\) \(\rightarrow\dfrac{98}{99}+\dfrac{143}{144}\)  

Giải:

\(A=\dfrac{2}{3}+\dfrac{14}{15}+\dfrac{34}{35}+\dfrac{62}{63}+\dfrac{98}{99}+\dfrac{143}{144}+\dfrac{194}{195}\) 

\(A=\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{35}\right)+...+\left(1-\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{195}\right)\) 

\(A=7-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{13.15}\right)\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{13.15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{15}\right)\right]\) 

\(A=7-\left[\dfrac{1}{2}.\dfrac{14}{15}\right]\) 

\(A=7-\dfrac{7}{15}\) 

\(A=\dfrac{98}{15}\) 

Chúc bạn học tốt!

A = \(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

A = ( 1 - 1/3 ) + ( 1 - 1/15 ) + ( 1 - 1/35 ) + ( 1 - 1/63 ) + ( 1 - 1/99 )

A = ( 1 + 1 + 1 + 1 + 1 ) - ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )

A = 5 - \(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

A  = 5 - ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 )

A = 5 - ( 1 - 1/11 ) 

A = 5 - 10/11

A = 45/11

Dấu \(.\)là dấu nhân 

\(A=\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}\)

\(\Rightarrow A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+\left(1-\frac{1}{35}\right)+\left(1-\frac{1}{63}\right)+\left(1-\frac{1}{99}\right)\)

\(\Rightarrow A=\left(1+1+1+1+1\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\)

\(\Rightarrow A=5-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\)

\(\Rightarrow A=5-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

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

\(\Rightarrow A=5-\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\Rightarrow A=5-\frac{1}{2}.\frac{10}{11}\)

\(\Rightarrow A=5-\frac{5}{11}\)

\(\Rightarrow A=\frac{55}{11}-\frac{5}{11}\)

\(\Rightarrow A=\frac{50}{11}\)

~ Ủng hộ nhé 

= 50/11 nha

Chúc bạn hok tốt nhé!