Cho \(A=1\times2+2\times3+3\times4+4\times5+...+100\times101\)
\(B=1\times3+2\times4+3\times5+4\times6+...+100\times102\)
Tính A-B?
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Ta có: B-A=1x3+2x4+3x5+4x6+...+100x102-(1x2+2x3+3x4+4x5+...+100x101)
=1x3+2x4+3x5+4x6+...100x102-1x2-2x3-3x4-4x5-...-100x101
=1+2+3+4+...+100
=((100-1):1+1)x((100-1):2)
=100x(101:2)
=5050
dau . la dau x
a/ 1.3.2.4.3.5.4.6.5.7/2.2.3.3.4.4.5.5.6.6=1.7/2.6=7/12
b/ ab.aba=abab
aba=abab:ab
aba=101
=>a=1 b=0
aabb : ab = 99 hay ab x 99 = aabb hay ab x100 – ab = aabb
Ta có phép tính
__ ab00
___ab___
aabb
b=0 hoặc b=5
Nếu b=0 thì a000 – a0 = aa00 (sai)
Nếu b=5 thì
__ a500
__a5___
aa55
a=4
c) thay a=7/6 b=6/5 thi 3 x a + 4 : b - 5/12=3.7/6+4.6/5-5/12=7/2+24/5-5/12=210/60+288/60-25/60=473/60
**** nha
\(\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(2.3.4.5.6\right).\left(3.4.5.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{7}{12}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
= 1 / 1 - 1 / 2 + 1 / 2 - 1 / 3 + 1 / 3 - 1 / 4 + 1 / 4 - 1 / 5 + 1 / 5 - 1 / 6
Ta gạch các ps trùng.
Còn lại :
1 / 1 - 1 / 6 = 6 / 5
a) \(\dfrac{30\times25\times7\times8}{75\times8\times12\times14}=\dfrac{3\times2\times5\times25\times7\times8}{25\times3\times8\times3\times4\times2\times7}=\dfrac{5}{3\times4}=\dfrac{5}{12}\)
b) \(\dfrac{8\times3\times4}{16\times3}=\dfrac{8\times3\times2\times2}{8\times2\times3}=2\)
c) \(\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{4\times5\times3\times2}{3\times5\times2\times4\times2}=\dfrac{1}{2}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=1\cdot3\cdot2\)
\(=6\)
\(A^2=6^2=36\)
Ta có: A=1.2+2.3+3.4+4.5+..............+100.101
B=1.3+2.4+3.5+4.6+...............+100.102
Vậy A-B=(1.2+2.3+3.4+4.5+..............+100.101)-(1.3+2.4+3.5+4.6+...............+100.102)
=(1.2-1.3)+(2.3-2.4)+(3.4-3.5)+(4.5-4.6)+..........+(100.101-100.102)
=(-1)+(-2)+(-3)+(-4)+..........+(-100)
=-(1+2+3+4+.........+100) có (100-1)+1=100 số hạng
=\(-\left[\left(100+1\right).100:2\right]\)
=-5050
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