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\(S=1\cdot2+2\cdot3+3\cdot4+...+1000\cdot1001\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+1000\cdot1001\cdot\left(1002-999\right)\)
\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+1000\cdot1001\cdot1002-999\cdot1000\cdot1001\)
\(3S=1000\cdot1001\cdot1002\Rightarrow S=\frac{1000\cdot1001\cdot1002}{3}=334.334.000\)
B = 1.2+2.3 +.......+1000.1001
3B= 1.2.3+2.3.4+3.4.3 +...... + 1000.1001.3
3B= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... .+ 1000.1001.(1002 - 999)
3B = (1.2.3 + 2.3.4 + 3.4.5 +...... + 1000.1001.1002) - (0.1.2 + 1.2.3 + 2.3.4 +.......+999.1000.1001)
3B = 1000.1001.1002 - 0.1.2
3B =1003002000
B = 334334000
B = 1.2+2.3 +.......+1000.1001
3B= 1.2.3+2.3.4+3.4.3 +...... + 1000.1001.3
3B= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... .+ 1000.1001.(1002 - 999)
3B = (1.2.3 + 2.3.4 + 3.4.5 +...... + 1000.1001.1002) - (0.1.2 + 1.2.3 + 2.3.4 +.......+999.1000.1001)
3B = 1000.1001.1002 - 0.1.2
3B =1003002000
B = 334334000
A =\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3\cdot4}+...+\frac{5}{99.100}\)
A = 5 x (\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) )
A = 5 x \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
A = 5 x \(\left(1-\frac{1}{100}\right)\)
A = 5 x \(\frac{99}{100}\)
A = \(\frac{495}{100}\)
A= \(\frac{99}{20}\)
Ta co : A =5.(1/1.2+1/2.3+1/3.4+....+1/99.100)
A= 5.(1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100)
Rut gon tung so ta co :A=5.(1-1/100)
A=5.99/100
A=1.99/50=99/50
\(\text{Ta có: }\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+.....+\frac{5}{99.100}\)
\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}\)
\(=\frac{99}{20}\)
5/1.2 + 5/2.3 + 5/3.4 + ... + 5/99.100
= 5 . ( 1/1.2 + 1/2.3 + 1/3.4 +... + 1/99.100 )
= 5 . ( 1 - 1/2 + 1/2 -1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 )
= 5 . ( 1 - 1/100 )
= 5 . 99/100
= 99/20
Xét: 3/(1.2)=(1+2)/(1.2)=1/2+1
5/(2.3)=(2+3)/(2.3)=1/3+/1/2
7/(3.4)=(3+4)/(3.4)=1/4+1/3
...
2021/(1010.1011)=(1010+1011)/(1010.1011)=1/1010+1/1011
Do đó: Q=1+1/2-1/2-1/3+1/3+1/4-...-1/1010-1/1011
=1-1/1011
=1010/1011
sai thì sorry nha, nhưng cách làm là đúng rồi
Lê Gia Long e lớp 4 thì ko đc lm!
\(Q=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}...-\frac{2021}{1010\cdot1011}\)
\(=\left(\frac{1}{1}+\frac{1}{2}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)...-\left(\frac{1}{1010}-\frac{1}{1011}\right)\)
\(=1-\frac{1}{1011}\)
\(=\frac{1010}{1011}\)
1) (x-3)(x-5) = 0
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
(x+7).35 = 2.35
\(\Rightarrow x+7=2\)
\(\Rightarrow x=2-7=-5\)
Vậy x = -5
2) 1.2 + 2.3 + 3.4 + .... + 99.100
Đặt A = 1.2 + 2.3 + .... + 99.100
3A = 1.2.3 + 2.3.4 + 3.4.3 + .... + 99.100.3
3A = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2)+ .... + 99.100.(101-98)
3A = ( 1.2.3 + 2.3.4 + 3.4.5 +.... + 99.100.101 ) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 98.99.100 )
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A = 999900
A = 999900 : 3
A = 333300
Vậy A = 333300
Mk ko chép lại đầu bài đâu,thông cảm nha mk chỉ biết giải ý B
3B=1.2.3+2.3.3+3.4.3+...+1000.1001.3
=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+1000.1001.(1002-999)
=1.2.3-1.2.0+2.3.4-2.3.1+3.4.5-3.4.2+...+1000.1001.1002-1000.1001.999
=(1.2.3+2.3.4+3.4.5+...+1000.1001.1002) - (1.2.0+2.3.1+3.4.2+...+1000.1001.999)
=1000.1001.1002
=>B=(1000.1001.1002):3
=334 334 000
k hộ mk nha!