11/10+41/40+89/88+...+461/460
Những câu hỏi liên quan
1.Tính nkanh
a, A= 1/10 + 1/40 + 1/88 + 1/154 + 1/238 + 1/340
b, 5/6 + 11/12 + 19/20 + 41/42 + 56/56 + 71/72 + 89/90
Mjnk đang cần gấp bn nào lm dc mik tick 7 tick nka
1. (-8)+12+(-20)+(-22)+28
2. 371+731+(-271)+(-531)
3.9+(-10)+11+(-12)+13+(-14)+15+(-16)
4.[461+(-78)+40]+(-461)
5.577+[(-100)+41]+(-618)
6.[453+64+(-879)]+(-517)
2/10+2/40+2/88+...+2/340+2/460=
2/10+2/40+2/88+...+2/340+2/460=
2/10+2/40+2/88+.....+2/340+2/460
=2/2.5+2/5.8+2/8.11+....+2/17.20+2/20.23
=2/3.(3/2.5+3/5.8+3/8.11+....+3/17.20+3/20.23)
=2/3.(1/2-1/5+1/5-1/8+1/8-1/11+.....+1/17-1/20+1/20-1/23)
=2/3.(1/2-1/23)=2/3.21/46=7/23
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2/10+2/40+2/88+...+2/340+2/460
tách thành
=2/2x5+2/5x8+2/8x11+...+2/17x20+2/20x23
=2/3x(1+1/5-1/5+1/8-1/8+...+1/20-1/20+1/23)
=2/3x(1+1/23)
=2/3x24/23
=16/23
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3/10+3/40+3/88+3/154+3/238+3/340+3/460
\(\frac{3}{10}+\frac{3}{40}+\frac{3}{88}+...+\frac{3}{460}\)
\(=\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{20\cdot23}\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{20}-\frac{1}{23}\)
\(=\frac{1}{2}-\frac{1}{23}\)
\(=\frac{21}{46}\)
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A =\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+.......+\frac{1}{460}\)
A=1/3x(1/2x5+1/5x8+......+1/20x23)
A=1/3x(1/2-1/5+1/5-1/8+......+1/20-1/23)
A=1/3x(1/2-1/23)
A=1/3x21/46
A=7/46
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7-8+9-10+11-12+...-86+87-88+89
\(=\left(7-8\right)+\left(9-10\right)+...+\left(87-88\right)+89\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+89\)
Số các số hạng từ 7 đến 88 là: \(88-7+1=82\) (số hạng)
\(=\left(-1\right)\cdot\left(82\div2\right)+89\)
\(=\left(-1\right)\cdot41+89\)
\(=\left(-41\right)+89\)
\(=48\)
cho so sánh hai phân số sau A= 11^89+1/11^90+1 và B=10^87+1/10^88+1
Sửa đề: B=11^87+1/11^88+1
\(11A=\dfrac{11^{90}+11}{11^{90}+1}=1+\dfrac{10}{11^{90}+1}\)
\(11B=\dfrac{11^{88}+11}{11^{88}+1}=1+\dfrac{10}{11^{88}+1}\)
mà 11^90>11^88
nên A<B
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