giải dùm đi tính S=4/5*7+4/7*9+...+4/59*61
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\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
4/ 5 * 7 + 4 / 7 * 9 + .. + 4 / 59 * 61
= 2/5 - 2/7 + 2/7 - 2/9 + ... + 2/59 - 2/61
= 2/5 - 2/61
= 112/305
Đặt
\(S=\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(S=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(S=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(S=2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(S=2.\frac{56}{305}=\frac{112}{305}\)
giải nhanh đi ko mik tiêu mất giải đúng cho 2 like
tính tổng
S=1+2+3-4-5-6+7+8+9-...+55+56+57-58-59-60
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
B = 1/4 x( 4/ 1x3x5 +4 /3x5x7 +...+ 4/47x49x51)
= 1/4 x( 1/ 1x3 - 1/3x5 +1/3x5 -1/5x7 +...+ 1/47x49 -1/49x51)
= 1/4 x( 1/1x3 -1/49x51)
= 1/4 x( 1/3 - 1/2499)
= 1/4 x 832/2499
= 208 /2499
Vậy B= 208 /2499
C= 1/ 9x11 -1/ 11x13 +1/ 11x13 -1/ 13x15 +... + 1/ 59x61 -1/61x63
= 1/ 9x11 -1/ 61x63
= 1/99 -1/3843
= 416 /42273
Vậy C= 416 /42273
\(A=\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{59\cdot61}=2\cdot\left(\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{59\cdot61}\right)\)
\(=2\cdot\left(\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)=2\cdot\left[\left(\frac{1}{7}-\frac{1}{61}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)+...+\left(\frac{1}{59}-\frac{1}{59}\right)\right]\)\(=2\cdot\left[\left(\frac{61}{427}-\frac{7}{427}\right)+0+...+0\right]=2\cdot\frac{54}{427}=\frac{108}{427}\)
Cách làm thì đúng rồi nhưng mk k chắc tính đúng, nhẩm thôi
xl bạn nhé nhưng bạn hiểu nhầm r ( chắc do mình ) nó là \(\frac{4\cdot9}{7}\)nhé bạn =))
\(A=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(A=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(A=\frac{84}{305}\)
\(\left(-\frac{3}{4}+\frac{59}{9}\right):\left(-\frac{7}{23}\right)+\left(\frac{5}{9}-\frac{1}{4}\right):\frac{7}{23}\)
= \(\left(-\frac{3}{4}+\frac{59}{9}\right).\left(-\frac{23}{7}\right)+\left(\frac{5}{9}-\frac{1}{4}\right).\left(\frac{23}{7}\right)\)
=\(\left(-\frac{23}{7}\right).\left(-\frac{3}{4}+\frac{59}{9}-\frac{5}{9}+\frac{1}{4}\right)\)
=\(\left(-\frac{23}{7}\right).\left(-\frac{2}{4}+\frac{54}{9}\right)=\left(-\frac{23}{7}\right).\left(-\frac{1}{4}+6\right)\)
=\(\left(-\frac{23}{7}\right).\frac{11}{2}=-\frac{253}{14}\)
1-3+5-7+9-...+57-59+61
=(1-3)+(4-7)+...+(57-59)+61
=(-2)+(-2)+...+(-2)+61
=(-2).30+61
=1
S=4/5.7+4/7.9+..+4/59.61
=>S=4/2.(2/5.7+2/7.9+...+2/59.61)
=>S=2.(1/5-1/7+1/7-1/9+...+1/59-1/61)
=>S=2.(1/5-1/61)=2.56/305=112/305
vậy S=112/305
Mot bai toan hay day