Mọi người đánh giúp mình nhé! Hạn là tối nay!!

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

a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)

\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)

\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)

\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)

\(=6:\left(-\frac{6}{12}\right)\)

\(=6:\left(-\frac{1}{2}\right)\)

\(=-12\)

b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )

=3/5: (-11/30) + 3/5 : (-7/5) 

=3/5:[-11/30+(-7/5)]

=3/5:53/30

=18/53

c) (1/2-13/14):5/7-(-2/21+1/7):5/7

= -3/7:5/7-1/21:5/7

=(-3/7-1/21):5/7

=-10/21:5/7

=-2/3

câu b vá c mình làm tắt nha. chúc bạn học tốt

A= (-4/5+4/3)+(-5/4+14/5)-7/3

= 8/15+31/20-7/3

= 25/12-7/3

= -1/4

B= 8/3.2/5.3/8.10.19/92

= 16/15.3/8.10.19/92

= 2/5.10.19/92

= 4.19/92

= 19/23

C= \(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+1\(\frac{5}{7}\)

=\(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)

= \(\frac{-10}{77}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)

= \(\frac{-10}{77}\)+\(\frac{-45}{98}\)+\(\frac{12}{7}\)

= \(\frac{-635}{1078}\)+\(\frac{12}{7}\)

= \(\frac{1213}{1078}\)

thanks nhìu

\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)

\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)

\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)

Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=0-66=-66\)

Auto làm nốt câu b

a,  Cộng cả 2 vế với 2 

Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)

\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)

=>  \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)

=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)

=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)

Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)

=> \(x=-65\)

b ,  Lm tương tự như Câu a

Chúc bn hok tốt

mình cần gấp nhé, xin cảm ơn

\(A=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{99}{5^{100}}\)

<=> \(5A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)

=> \(5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^5}+...+\frac{99}{5^{100}}\right)\)

=> \(4A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

=> \(20A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\)

=> \(20A-4A=\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{99}}-\frac{99}{5^{100}}\right)\)

=> \(16A=1+\frac{99}{5^{100}}-\frac{100}{5^{99}}< 1\)

=> \(A< \frac{1}{16}< \frac{1}{6}\)

\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)

\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)

\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

\(=\frac{5}{7}.0=0\)

c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)

\(=\frac{27}{5}.10=27.2=54\)

\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)

\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)