Câu 1.

$\frac{1}{15}-\frac{9}{15}=\frac{-8}{15}$

$\frac{2}{15}-\frac{10}{15}=\frac{-8}{15}$

$\frac{3}{15}-\frac{11}{15}=\frac{-8}{15}$

Câu 2:

$\frac{-9}{15}+\frac{1}{15}=\frac{-8}{15}$

$\frac{-10}{15}+\frac{2}{15}=\frac{-8}{15}$

$\frac{-11}{15}+\frac{3}{15}=\frac{-8}{15}$

Câu 3:

$\frac{-7}{15}-\frac{1}{15}=\frac{-8}{15}$

$\frac{-6}{15}-\frac{2}{15}=\frac{-8}{15}$

$\frac{-5}{15}-\frac{3}{15}=\frac{-8}{15}$

giúp mình với mình cần gấp lắm

1/ Cách 1: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-1+\left(-7\right)}{15}\) = \(\dfrac{-1}{15}\) + \(\dfrac{-7}{15}\)

Cách 2: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-2+\left(-6\right)}{15}\) = \(\dfrac{-2}{15}\) + \(\dfrac{-6}{15}\) = \(\dfrac{-2}{15}\) + \(\dfrac{-2}{5}\)

Cách 3: Ta có: \(\dfrac{-8}{15}\) = \(\dfrac{-3+\left(-5\right)}{15}\) = \(\dfrac{-3}{15}\) + \(\dfrac{-5}{15}\) = \(\dfrac{-1}{5}\) + \(\dfrac{-1}{3}\)

2/ C1: Ta có: \(\dfrac{-8}{15}=\dfrac{10-18}{15}=\dfrac{10}{15}-\dfrac{18}{15}=\dfrac{2}{3}-\dfrac{6}{5}\)

C2: Ta có: \(\dfrac{-8}{15}=\dfrac{1-9}{15}=\dfrac{1}{15}-\dfrac{9}{15}=\dfrac{1}{15}-\dfrac{3}{5}\)

C3: Ta có: \(\dfrac{-8}{15}=\dfrac{5-13}{15}=\dfrac{5}{15}-\dfrac{13}{15}=\dfrac{1}{3}-\dfrac{13}{15}\)

3/C1: Ta có: \(\dfrac{-8}{15}=\dfrac{-16+8}{15}=\dfrac{-16}{15}+\dfrac{8}{15}\)

C2: Ta có: \(\dfrac{-8}{15}=\dfrac{-20+12}{15}=\dfrac{-20}{15}+\dfrac{12}{15}=\dfrac{-4}{3}+\dfrac{4}{5}\)

C3: Ta có:\(\dfrac{-8}{15}=\dfrac{-14+6}{15}=\dfrac{-14}{15}+\dfrac{6}{15}=\dfrac{-14}{15}+\dfrac{2}{5}\)

Mk chỉ nghĩ z thôi chứ ko biết đúng hay sai nữa, có j thì góp ý nha. Chúc bn hc tốt!!!

a: \(\dfrac{-8}{15}=\dfrac{-2}{15}+\dfrac{-6}{15}\)

\(\dfrac{-8}{15}=\dfrac{-1}{15}+\dfrac{-7}{15}\)

\(\dfrac{-8}{15}=\dfrac{-4}{15}+\dfrac{-4}{15}\)

b: \(-\dfrac{8}{15}=\dfrac{17}{15}-\dfrac{25}{15}\)

\(\dfrac{-8}{15}=\dfrac{9}{15}-\dfrac{17}{15}\)

\(\dfrac{-8}{15}=\dfrac{10}{15}-\dfrac{18}{15}\)

\(\frac{-2}{15}\) + \(\frac{2}{15}\)

\(\frac{-3}{15}\) + \(\frac{3}{15}\)

\(\frac{-4}{15}\)+ \(\frac{4}{15}\)

Mình làm câu a

\(Để\frac{a}{b}< \frac{a+c}{b+d}\) thì a(b+d) < b(a+c) ↔ ab + ad , ab + bc ↔ ab < bc ↔ \(\frac{a}{b}< \frac{c}{d}\)

\(Để\frac{a+c}{b+d}< \frac{c}{d}\) thì (a+c).d < (b+d).c ↔ ad + cd < bc + cd ↔ ab < bc ↔ \(\frac{a}{b}< \frac{c}{d}\)

nhân chéo thôi

3.- Xét a(b+2001)=ab+2001a

        b(a+2001)=ab+2001b

- Ta xét 3 trường hợp sau:

+Nếu a>b =>2001a>2001b

                 =>a(b+2001)>b+(a+2001)

                 =>a/b > a+2001/b+2001

+Nếu a<b =>2001a<2001b

                 =>a(b+2001)<b+(a+2001)

                 =>a/b < a+2001/b+2001

+Nếu a=b =>a/b = a+2001/b+2001

a)\(\dfrac{-3}{15}+\dfrac{-5}{15}\)

b)\(\dfrac{4}{3}-\dfrac{4}{5}\)

c)\(-1+\dfrac{7}{15}\)

d)\(\dfrac{-1}{15}-\dfrac{7}{15}\)

a: \(\dfrac{-8}{15}=\dfrac{-6}{15}+\dfrac{-2}{15}\)

\(\dfrac{-8}{15}=\dfrac{-3}{15}+\dfrac{-5}{15}\)

\(\dfrac{-8}{15}=\dfrac{-1}{15}+\dfrac{-7}{15}\)

b: \(\dfrac{-8}{15}=\dfrac{16}{15}-\dfrac{24}{15}\)

\(\dfrac{-8}{15}=\dfrac{22}{15}-\dfrac{30}{15}\)

\(\dfrac{-8}{15}=\dfrac{9}{15}-\dfrac{17}{15}\)