\(\left(-\frac{7}{24}-0,75+25\%\right):\left(-2\frac{1}{8}\right)\)   

\(=\left(-\frac{7}{24}-\frac{18}{24}+\frac{6}{24}\right):\left(-\frac{17}{8}\right)\) 

\(=\left(-\frac{19}{24}\right)\cdot\left(-\frac{8}{17}\right)\)  

\(=\frac{19}{51}\)

A=1+2/20 mũ 18 -1

b=1+2/20 mũ 18 -3

vì 1=1 so sánh 2/20 mũ 18-1 và 20 mũ 18 -3

a<b

Vì a;b;c > 0 => \(\frac{a}{c+a}< \frac{a+b}{a+b+c}\) ; 

\(\frac{b}{a+b}< \frac{b+c}{a+b+c}\);

\(\frac{c}{b+c}< \frac{a+c}{a+b+c}\)

Khi đó \(\frac{a}{c+a}+\frac{b}{a+b}+\frac{c}{b+c}< \frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{c+a}{a+b+c}\)

\(=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)(đpcm)

Vì a;b;c > 0 nên\(\frac{a}{a+b}>\frac{a}{a+b+c}\);

\(\frac{b}{b+c}>\frac{b}{a+b+c}\);

\(\frac{c}{c+a}>\frac{c}{a+b+c}\)

Khi đó \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)(đpcm)

\(\frac{1}{2.5}\)\(+\)\(\frac{1}{5.8}\)\(+\frac{1}{8.11}\)\(+...+\frac{1}{152.155}\)

=\(\frac{1}{2}\) \(-\frac{1}{5}\) \(+\frac{1}{5}\) \(-\frac{1}{8}\) \(+...+\frac{1}{152}\) \(-\frac{1}{155}\)

=\(\frac{1}{2}\)\(-\frac{1}{155}\)

=\(\frac{153}{310}\)

a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{1000}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)...\left(-\frac{999}{1000}\right)\)

\(=-\frac{1.2.3...999}{2.3.4...1000}=-\frac{1}{1000}\)

b)\(B=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}:\frac{3}{4}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}:\frac{3}{4}=\frac{3}{4}:\frac{3}{4}=1\)

d) \(D=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}+\frac{1}{1024}\)

=> \(2D=2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\)

=> \(2D-D=\left(2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)

=> \(D=2-\frac{1}{1024}=\frac{2047}{1024}\)

bằng 34 bạn nhé

=83 độ 

nha bn

Học tốt

Trên cùng một bờ mặt phẳng chứa tia Oi, ta có \(\widehat{KOI}< \widehat{MOI}\left(49^o< 132^o\right)\)nên tia Ok nằm giữa hai tia Oi và Om

\(\widehat{KOI}+\widehat{KOM}=\widehat{MOI}\)

\(hay49^o+\widehat{KOM}=132^o\)

\(\widehat{KOM}=132^o-49^o\)

\(\widehat{KOM}=83^o\)

Câu 3. Đúng ghi Đ, sai ghi S.

a) 2,3 giờ < 2 giờ 30 phút   s       

  c) 125 x 0,8 > 100 s

b) 3 m2 6 dm2 = 3,06 m2     đ     

   d) 5 kg = 0,05 tạ đ

\(a,1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)

\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right]\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)

                         \(10\)chữ số

\(=\left(-1\right).10\)

\(=-10\)

\(a\text{)}1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)

\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right]\)

\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\left(\text{có 10 số hạng -1}\right)\)

\(=\left(-1\right).10=-10\)