\(1,|a-b+c|=|-8+5+6|=|3|=0\)

\(2,|a+b-c|=|-8-5-6|=|-19|=19\)

\(3,|a-b-c|=|-8+5-6|=|-9|=9\)

\(4,|-a+b+c|=|8-5+6|=|9|=9\)

Đặt 

\(A=\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+..+\frac{1}{2011}}=\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2012}}{1+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2012}\right)}=\frac{1}{2012}\)

Ta có 10x=5y=6z

=>y=2x

=>z=5x/3

=>x+y+z=x+2x+5x/3=14x/3=56

=>x=12

=>y=24,z=20

\(a.\frac{\left(1+2+..+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right)\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}}=\frac{\left(1+2+..+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{100}}=0\)\(b.\frac{-\frac{6}{7}+\frac{6}{19}-\frac{6}{31}}{\frac{9}{7}-\frac{9}{19}+\frac{9}{31}}=-\frac{6\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}{9\left(\frac{1}{7}-\frac{1}{19}+\frac{1}{31}\right)}=-\frac{6}{9}=-\frac{2}{3}\)

\(c.\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

 \(-\frac{8}{18}\)\(-\frac{15}{27}\)

\(=-\frac{4}{9}\)\(-\frac{5}{9}\)

\(=-\frac{4}{9}\)\(+-\frac{5}{9}\)

\(=-\frac{9}{9}\)\(=-1\)

Học tốt nha em

\(-\frac{8}{18}-\frac{15}{37}=-\frac{4}{9}-\frac{15}{37}=-\frac{148+135}{333}=-\frac{283}{333}\)

 A = x+1/x-2
<=> A = x-2+3/x-2 = 1 + 3/x-2
Để x nguyên thì 3/x-2 nguyên (vì 1 nguyên)
-> x-2 thuộc Ư(3)
-> x-2 thuộc {-3,-1,1,3}
-> x thuộc {-1; 1; 3; 5}

B=Để 2x-1/x+5 là số nguyên thì

2x - 1 ⋮ x + 5

=> (2x - 1) - [2(x + 5)] ⋮ x + 5

=> (2x - 1) - (2x + 10) ⋮ x + 5

=> 2x - 1 - 2x - 10 ⋮ x + 5

=> 9 ⋮ x + 5

=> x + 5 ∈ Ư₍₉₎ = {1; 3; 9; -1; -3; -9}

=> x ∈ {-4; -2; 4; -6; -8; -14}

a, x = -1 ; 1 ; 3 ; 5

b, x = 4 ; -6 ; 6 ; -16

c, x = 0 ; 1 ; 2 ; 3

xin tiick

\(2^{600}=\left(2^3\right)^{200}=8^{200}\)

\(3^{400}=\left(3^2\right)^{200}=9^{200}\)

VÌ \(8^{200}< 9^{200}\)

---> \(2^{600}< 3^{400}\)

XIN TIICK

2 mũ 600 nhé mà a zai lớp 6 hỏi cái này làm gì 
t i c k e nhé a zai :)