tính :
a. 3 + 2\5 b. a - 5\7 c. 1- < 2\5 + 1\3 >
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C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{100}-\left(\frac{1}{2.1}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
=\(\frac{1}{100}-\frac{99}{100}\)
=\(\frac{-98}{100}=\frac{-49}{50}\)
C=1/100 -1/100.99 -1/99.98 -1/98.97-......- 1/3.2 -1/2.1
= 1/100 - (1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1)
Đặt A = 1/100.99 + 1/99.98 + 1/98.97-......+ 1/3.2 +1/2.1 => C = 1/100 - A
Dễ thấy 1/2.1 = 1/1 - 1/2
1/3.2 = 1/2 - 1/3
.....................
1/99.98 = 1/98 - 1/99
1/100.99 = 1/99 - 1/100
=> cộng từng vế với vế ta
a) \(2^5+5.13-3.2^3\)
\(=32+5.13-3.8\)
\(=32+65-24\)
\(=97-24\)
\(=73\)
b) \(5^{13}:5^{10}-5^2.2^2\)
\(=5^3-25.4\)
\(=125-100\)
\(=25\)
c) \(4^5:4^3-3^9:3^7+5^0\)
\(=4^2-3^2+1\)
\(=16-9+1\)
\(=7+1\)
\(=8\)
a)\(\frac{2}{3}\times\frac{4}{5}-\frac{3}{5}\times\frac{1}{2}=\frac{8}{15}-\frac{3}{10}\)
\(=\frac{7}{30}\)
b) \(\frac{12}{5}-\frac{4}{7}-\frac{3}{7}=\frac{64}{35}-\frac{3}{7}\)
\(=\frac{7}{5}\)
cho chj nhé em THKS em nhìu!
CHÚC EM HỌC TỐT!
a)1/2-7/13-1/3+ -6/12+1/2+1 1/3
=(1/2-1/2)+(1 1/3 - 1/3)-6/12+7/13
=0+1-1/2+7/13
=1-1/2+7/13
=1/2+7/13
=27/26
b)0,75+2/5+1/9 - 1 2/5 + 5/4
=3/4+2/5+1/9-1 2/5+5/4
=(3/4+5/4)-(1 2/5-2/5)+1/9
=2-1+1/9
=1+1/9
=10/9
c) (-1 1/2:3/4).(-4 1/2)-1/2
=(-1,5:0,75).(-4,5)-0,5
=-2.-4,5-0,5
=9-0,5
=8,5
a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)
\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)
\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)
\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)
\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
A = 0-1 + 2-3 + 4-5 +...+ 2017-2018
=> A = (-1) + (-1) + (-1) +...+ (-1) (Có 1009 số hạng)
=> A = 1009.(-1)
=> A = -1009
B = 1-3+5-7+ 9-11+....+2005-2007
=> B = (-2) + (-2) +(-2) +...+ (-2) (Có 502 số hạng)
=> B = 502.(-2)
=> B = -1004
C=1+2+3-4-5-6+7+8+9-10-11-12+.....+97+98+99-100-101-102
=> C = (1+2+3-4-5-6)+...+(97+98+99-100-101-102) (có 17 cặp số)
=> C = (-9) + (-9) +...+ (-9) (có 17 số hạng)
=> C = (-9).17
=> C = -153
Bài 1: \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\) (1)
Từ \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\)
Thay vào (1) ta có:
\(\frac{a^2+ab}{b^2+ab}=\frac{a}{b}\Rightarrow\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\) (luôn đúng)
Vậy ta có điều phải chứng minh
Bài 2:
\(B=\dfrac{2^{15}\cdot5^8-2^5\cdot2^9\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{5}{4}\cdot\dfrac{-3}{6}=\dfrac{5}{4}\cdot\dfrac{-1}{2}=-\dfrac{5}{8}\)
a.3+2/5
=15/5+2/5
=17/5
b.4-5/7
=28/7-5/7
=23/7
c.1-(2/5+1/3)
=1-11/15
=4/15
a)3+2/5=17/5
b) a-5/7=17/5-5/7=94/35
c)1-(2/5+1/3)=4/15