P = \(2\times3+3\times4+4\times5+5\times6+...+99\times100\)
Tính nhanh: D=\(\frac{4}{2\times3\times4}+\frac{4}{3\times4\times5}+\frac{4}{4\times5\times6}+.......+\frac{4}{98\times99\times100}\)
Tính nhanh:
C=\(\frac{3}{2\times3\times4}+\frac{3}{3\times4\times5}+\frac{3}{4\times5\times6}+......+\frac{3}{98\times99\times100}\)
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+.....+\frac{3}{98.99.100}\)
C = \(3.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{99}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
Hồ Thu Giang cần chứ nếu được cảm ơn nha
CHO A = \(\frac{2\times9\times8+3\times12\times10+4\times15\times12+...+98\times297\times200}{2\times3\times4+3\times4\times5+4\times5\times6+...+98\times99\times100}\)
TÍNH A\(^2\)
\(A=\frac{2\cdot9\cdot8+3\cdot12\cdot10+4\cdot15\cdot12+...+98\cdot297\cdot200}{2\cdot3\cdot4+3\cdot4\cdot5+4\cdot5\cdot6+...+98\cdot99\cdot100}\)
\(=\frac{2\cdot1\cdot3\cdot3\cdot4\cdot2+3\cdot1\cdot4\cdot3\cdot5\cdot2+...+98\cdot1+99\cdot3+100\cdot2}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=\frac{1\cdot3\cdot2\cdot\left(2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\right)}{2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100}\)
\(=1\cdot3\cdot2\)
\(=6\)
\(A^2=6^2=36\)
a)\(\frac{1\times3\times2\times4\times3\times5\times4\times6\times5\times7}{2\times2\times3\times3\times4\times4\times5\times5\times6\times6}=?\)
b)Thay chữ a, b bằng chữ số thích hợp:
1) ab x aba = abab 2) 99 x ab = aabb
c) 3 x a + 4 : b - 5/12 với a = 7/6, b = 6/5
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
\(\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}=\frac{\left(2.3.4.5.6\right).\left(3.4.5.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{7}{12}\)
chỉ trả lời được câu a thôi: phần a bằng 7/12
\(\frac{1\times2}{2\times3}+\frac{2\times3}{3\times4}+\frac{3\times4}{4\times5}+...+\frac{98\times99}{99\times100}\)
\(=\frac{1.2}{99.100}\)
\(=\frac{2}{9900}=\frac{1}{4950}\)
tính nhanh: \(\frac{1\times3\times2\times4\times3\times5\times4\times6\times5\times7}{2\times2\times3\times3\times4\times4\times5\times5\times6\times6}\)
tính tổng \(S=1\times2+2\times3+3\times4+4\times5....+99\times100\)
ta có :
S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101
S = 99x100x101 : 3
S = 333300
tính :\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+\frac{1}{4\times5\times6}+\frac{1}{5\times6\times7}+\frac{1}{6\times7\times8}+\frac{1}{7\times8\times9}+\frac{1}{8\times9\times10}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\frac{1\times2\times3\times4\times5}{2\times3\times4\times5\times6}\)