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\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
\(\Rightarrow A=\frac{5}{11}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow B=\frac{1009}{2019}\)
\(\frac{2}{7}C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2017.2019}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\Rightarrow C=\frac{2018}{2019}:\frac{2}{7}=\frac{7063}{2019}\)
1 + 2 -3-4 + 5 + 6-7-8+...+2017+2018
= 1 + (2-3-4+5) + (6-7-8+9) + ...+ (2014-2015-2016+2017) + 2018
= 1 + 0+0+...+0+2018
=2 019
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)
A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)
A=3.(1-1/400)
A=3.399/400
A=1197/400
\(\left(x-7\right)\left(x+2019\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+2019=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2019\end{cases}}\)
\(9-25=\left(7-x\right)-\left(25+7\right)\)
\(\Leftrightarrow-16=7-x-25-7\)
\(\Leftrightarrow-x=-16+25\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
\(2\left(4x-2x\right)-7x=15\)
\(\Leftrightarrow4x-7x=15\)
\(\Leftrightarrow x=-5\)
a ) 9 - 25 = ( 7 - x ) - ( 25 + 7 )
9 - 25 = 7 - x - 25 - 7
9 - 25 - 7 + 25 + 7 = -x
9 = - x
=> x = -9
Vậy x = -9
b) 2 . ( 4x - 2x ) - 7x = 15
8x - 4x - 7x = 15
-3x = 15
x = 15 : ( - 3 )
x = -5
Vậy x = -5
c ) ( x - 7 ). ( x + 2019 ) = 0
=> x - 7 = 0 hoặc x + 2019 = 0
=> x = 7 hoặc x = - 2019
vậy x \(\in\){ 7 ; -2019 }
1.S1=1 - 2 + 3 - 4 + ... + 1997 - 1998 + 1999
= (1 - 2) + ...+(1997 - 1998) + 1999
= -1 + -1 + ...+-1 + 1999
SH:1998 : 2
= 999 . -1
= -999
TDS:-999 + 1999
= 1000
b.S2=1 - 4 + 7 - 10 + ...- 2998+3001
= (1 - 4) + (7 - 10) + ...+ (2995 - 2998) + 3001
= -3 + -3 + ...+-3 + 3001
= (2998 - 1) : 3 + 1
= 1000 . -3
= -3000 + 3001
= 1
câu b mình làm lộn :
S2=1000 : 2
= 500 . -3
=-1500 + 3001
= 1501
KẾT QUẢ RA 1501 NHA
Bài 1: Tính nhanh:
A = 3/1*2 + 3/2*3 + 3/3*4 + ... + 3/399*400
=>3A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/399*400
3A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/399 - 1/400
3A = 1 - 1/400
3A = 400/400 - 1/400
3A = 399/400
A = 399/400 : 3
A = 399/400 . 1/3
A = 133/400.
Có gì ko hiểu bn ib mk nha.^^
\(A=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{399.400}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(A=3.\left(1-\frac{1}{400}\right)\)
\(A=3.\frac{399}{400}\)
\(A=\frac{1197}{400}\)
\(B=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{399.400}\)
\(B=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)
\(B=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)
\(B=5.\left(1-\frac{1}{400}\right)\)
\(B=5.\frac{399}{400}\)
\(B=\frac{399}{80}\)
\(C=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\)
\(C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\)
\(C=\frac{1}{5}-\frac{1}{151}\)
\(C=\frac{146}{755}\)
\(D=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{149.151}\)
\(D=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{151}\right)\)
\(D=\frac{3}{2}.\frac{146}{755}\)
\(D=\frac{219}{755}\)
\(E=\frac{11}{1.3}+\frac{11}{3.5}+\frac{11}{5.7}+...+\frac{11}{99.101}\)
\(E=\frac{11}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\left(1-\frac{1}{101}\right)\)
\(E=\frac{11}{2}.\frac{100}{101}\)
\(E=\frac{550}{101}\)
_Chúc bạn học tốt_
A = SCSH: ( 102 - 1 ) : 1 + 1 = 102
A = Tổng: ( 102 + 1 ) . 102 : 2 = 5253
Vậy KQ là: 5253
B = SCSH: ( 2998 - 1 ) : 3 + 1 = 1000
B = Tổng: ( 2998 + 1 ) . 1000 : 2 = 1499500
Vậy KQ là 1499500
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7.9}+\cdot\cdot\cdot\cdot\cdot+\frac{1}{43\cdot45}\)
=\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\cdot\cdot\cdot\cdot\cdot+\frac{1}{43}-\frac{1}{45}\right)\)
=\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{45}\right)\)
=\(\frac{1}{2}\cdot\frac{14}{45}\)
=\(\frac{7}{45}\)
Cảm ơn bạn nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nhiều nha