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Ta thấy:
1 x 4 = 1 x 2 + 1 x 2
2 x 5 = 2 x 3 + 2 x 2
3 x 6 = 3 x 4 + 3 x 2
.................................
Suy ra:
D = (1 x 2 + 2 x 3 + 3 x 4 + .... + 97 x 98) + (1 x 2 + 2 x 2 + 3 x 2 + .... + 97 x 2)
D = (1x2+2x3+3x4+...+97x98) + (1+2+3+...+99)x2
D = (1x2+2x3+3x4+...+97x98) + 100 x 99 : 2
D - 100 x 99 : 2 = 1x2+2x3+3x4+...+97x98
D - 4950 = 1x2+2x3+3x4+...+97x98
(D - 4950) x 3 = 1x2x(3-0)+2x3x(4-1)+3x4x(5-2)+......+97x98x(99-96)
(D-4950)x3 = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + .... + 97 x 98 x 99 - 96 x 97 x 98
(D-4950)x3 = 97 x 98 x 99
Và từ đây ta có thể tìm hướng để ra kết quả
\(A=20\times21+21\times22+...+99\times100\)
\(3\times A=20\times21\times\left(22-19\right)+21\times22\times\left(23-20\right)+...+99\times100\times\left(101-98\right)\)
\(=20\times21\times22-19\times20\times21+...+99\times100\times101-98\times99\times100\)
\(=99\times100\times101-19\times20\times21\)
Suy ra \(A=\frac{99\times100\times101-19\times20\times21}{3}=360640\)
\(B=3\times4\times5+4\times5\times6+...+98\times99\times100\)
\(4\times B=3\times4\times5\times\left(6-2\right)+4\times5\times6\times\left(7-3\right)+...+98\times99\times100\times\left(101-97\right)\)
\(=3\times4\times5\times6-2\times3\times4\times5+...+98\times99\times100\times101-97\times98\times99\times100\)
\(=98\times99\times100\times101-2\times3\times4\times5\)
Suy ra \(B=\frac{98\times99\times100\times101-2\times3\times4\times5}{4}=24497520\)
Bài 1:
a: Tổng là:
(-19+19)+(-18+18)+...+20=20
b: Tổng là:
-18+(-17+17)+...+0=-18
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
\(\text{1: 1+(-2)+3+(-4)+........+19+(-20)}\)
\(\text{=1-2+3-4+.....+19-20}\)
\(\text{=-1+(-1)+..........+(-1)}\)
\(\text{=-1.(20-1)+1:2}\)
\(\text{=-1.10}\)
\(=-10\)
\(\text{2: 1- 2 +3 -4+......+99 - 100}\)
\(\text{=( 1- 2) +(3 -4)+......+(99 - 100)}\)
\(\text{=-1+(-1)+.....+(-1)}\)
\(\text{=-1.(100-1)+1:2}\\ \text{=-1.50}\)
\(=-50\)
\(\text{3: 2 - 4+6 - 8+.....+48 - 50}\\ \text{=(2 - 4)+(6 - 8)+.....+(48 - 50)}\\ \text{=-2+(-2)+......+(-2)}\\ \text{=-2.(50-2):2+1}\\ \text{=-2.25}\\ =-50\)
\(\text{4: -1 + 3 - 5 +7 -.....+97 - 99}\\ \text{=-1 +(3-5)+(7-9)+.......+(97-99)}\\ \text{=-1+(-2)+(-2)+........+(-2)}\\ \text{=-1+[(-2).(99-3):2+1]}\\ \text{=-1+[(-2).49]}\\ \text{=-1+(-98)}\\ =-99\)
\(\text{5: 1 + 2 – 3 – 4 + . . . . + 97 + 98 – 99 – 100}\)
\(\text{=1−2+3−4+5−6+...+99−100}\)
\(\text{=(1−2)+(3−4)+(5−6)+...+(99−100) }\)
\(\text{=(−1)+(−1)+(−1)+...+(−1)}\)
\(\text{=(−1).50}\)
\(\text{=−50}\)
1/ 1 + (-2) + 3 + (-4) + . . . + 19 + (-20)
=(1-2)+(3-4)+..........+(19-20)
=-1+(-1)+.......+(-1)
=-1.[(20-1)+1:2]
=-1.10
=-10
2/ 1 – 2 + 3 – 4 + . . . + 99 – 100
=-1+(-1)+....+(-1)
=-1.[(100-1)+1:2]
=-1.50
=-50
3/ 2 – 4 + 6 – 8 + . . . + 48 – 50
=-2+(-2)+.......+(-2)
=-2.[(50-2):2]:2
=-2.12
=-24
=-2.
Ta có :
\(\frac{x+1}{100}+\frac{x+2}{99}=\frac{x+3}{98}+\frac{x+4}{97}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{100}+1\right)+\left(\frac{x+2}{99}+1\right)=\left(\frac{x+3}{98}+1\right)+\left(\frac{x+4}{97}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}=\frac{x+101}{98}+\frac{x+101}{97}\)
\(\Leftrightarrow\)\(\frac{x+101}{100}+\frac{x+101}{99}-\frac{x+101}{98}-\frac{x+101}{97}=0\)
\(\Leftrightarrow\)\(\left(x+101\right)\left(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\right)=0\)
Vì \(\frac{1}{100}+\frac{1}{99}-\frac{1}{98}-\frac{1}{97}\ne0\)
Nên \(x+101=0\)
\(\Rightarrow\)\(x=-101\)
Vậy \(x=-101\)
Chúc bạn học tốt ~
b)Tận cùng=5 hoặc 0 nhưng mình ngại viết lắm,thông cảm nha
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