a: \(\dfrac{-7}{6}=\dfrac{-7\cdot3}{6\cdot3}=\dfrac{-21}{18}\)

\(\dfrac{-11}{9}=\dfrac{-11\cdot2}{9\cdot2}=\dfrac{-22}{18}\)

mà -21>-22

nên \(-\dfrac{7}{6}>-\dfrac{11}{9}\)

b: \(\dfrac{5}{-7}=\dfrac{-5}{7}=\dfrac{-5\cdot5}{7\cdot5}=\dfrac{-25}{35}\)

\(\dfrac{-4}{5}=\dfrac{-4\cdot7}{5\cdot7}=\dfrac{-28}{35}\)

mà -25>-28

nên \(\dfrac{5}{-7}>\dfrac{-4}{5}\)

c: \(\dfrac{-8}{7}< -1\)

\(-1< -\dfrac{2}{5}\)

Do đó: \(-\dfrac{8}{7}< -\dfrac{2}{5}\)

d: \(-\dfrac{2}{5}< 0\)

\(0< \dfrac{1}{3}\)

Do đó: \(-\dfrac{2}{5}< \dfrac{1}{3}\)

Help me pls

a) \(\dfrac{5}{9}< \dfrac{7}{9}\)

b) \(\dfrac{7}{6}>\dfrac{6}{6}\)

c) \(\dfrac{3}{14}< \dfrac{5}{14}\)

d) \(\dfrac{5}{8}< \dfrac{9}{8}\)

a) \(\dfrac{2}{5}=\dfrac{4}{10}\)

\(\dfrac{4}{10}>\dfrac{3}{10}\)

b) \(\dfrac{5}{6}=\dfrac{10}{12}\)

\(\dfrac{7}{12}< \dfrac{10}{12}\)

c) \(\dfrac{1}{2}=\dfrac{2}{4}\)

\(\dfrac{3}{4}< \dfrac{2}{4}\)

d) \(\dfrac{8}{3}=\dfrac{56}{21}\)

\(\dfrac{56}{21}>\dfrac{11}{21}\)

Giải:

a)Ta có:

C=1957/2007=1957+50-50/2007

                      =2007-50/2007

                      =2007/2007-50/2007

                      =1-50/2007

D=1935/1985=1935+50-50/1985

                      =1985-50/1985

                      =1985/1985-50/1985

                      =1-50/1985

Vì 50/2007<50/1985 nên -50/2007>-50/1985

⇒C>D

b)Ta có:

A=2016²⁰¹⁶+2/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1/2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=1+3/2016²⁰¹⁶-1

Tương tự: B=2016²⁰¹⁶/2016²⁰¹⁶-3

                 B=1+3/2016²⁰¹⁶-3

Vì 2016²⁰¹⁶-1>2016²⁰¹⁶-3 nên 3/2016²⁰¹⁶-1<3/2016²⁰¹⁶-3

⇒A<B

Chúc bạn học tốt!

Làm tiếp:

c)Ta có:

M=10²⁰¹⁸+1/10²⁰¹⁹+1

10M=10.(10²⁰¹⁸+1)/20²⁰¹⁹+1

10M=10²⁰¹⁹+10/10²⁰¹⁹+1

10M=10²⁰¹⁹+1+9/10²⁰¹⁹+1

10M=10²⁰¹⁹+1/10²⁰¹⁹+1 + 9/10²⁰¹⁹+1

10M=1+9/10²⁰¹⁹+1

Tương tự:

N=10²⁰¹⁹+1/10²⁰²⁰+1

10N=1+9/10²⁰²⁰+1

Vì 9/10²⁰¹⁹+1>9/10²⁰²⁰+1 nên 10M>10N

⇒M>N

Chúc bạn học tốt!

con cặc

a) \(\dfrac{23}{24}< 1\)

\(\dfrac{24}{23}>1\)

\(\Rightarrow\dfrac{23}{24}< \dfrac{24}{23}\)

b) \(\dfrac{4}{21}< \dfrac{4}{20}=\dfrac{1}{5}=\dfrac{6}{30}< \dfrac{6}{29}\)

c) \(\dfrac{6}{7}=1-\dfrac{1}{7}< \dfrac{8}{9}=1-\dfrac{1}{9}\)

d) \(\dfrac{1212}{1313}=\dfrac{12\times101}{13\times101}=\dfrac{12}{13}\)

a)

b)

+) Quy đồng mẫu số ba phân số $\frac{1}{4};\frac{3}{4};\frac{5}{8}$

$\frac{1}{4} = \frac{{1 \times 2}}{{4 \times 2}} = \frac{2}{8}$

$\frac{3}{4} = \frac{{3 \times 2}}{{4 \times 2}} = \frac{6}{8}$ ; Giữ nguyên phân số $\frac{5}{8}$

Vì $\frac{2}{8} < \frac{5}{8} < \frac{6}{8}$ nên $\frac{1}{4} < \frac{5}{8} < \frac{3}{4}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là: $\frac{1}{4};\,\,\frac{5}{8};\,\,\frac{3}{4}$

+) Quy đồng mẫu số ba phân số $\frac{2}{3};\,\,\frac{2}{9};\,\,\frac{5}{9}$

$\frac{2}{3} = \frac{{2 \times 3}}{{3 \times 3}} = \frac{6}{9}$ ; Giữ nguyên phân số $\frac{2}{9}$; $\frac{5}{9}$

Vì $\frac{2}{9} < \frac{5}{9} < \frac{6}{9}$ nên $\frac{2}{9} < \frac{5}{9} < \frac{2}{3}$

Vậy các phân số xếp theo thứ tự từ bé đến lớn là $\frac{2}{9};\,\,\frac{5}{9};\,\,\frac{2}{3}$

2/

a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)

\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)

-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)

b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)

-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)

c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)

-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)

a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)

c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)

a) 

b) 

c) \(\dfrac{25}{100}=\dfrac{10}{40}\)

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

\(x=-\dfrac{15}{2}\)

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

\(x=\dfrac{-1}{2}:\dfrac{3}{4}\)

\(x=\)\(\dfrac{-2}{3}\)

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

\(\dfrac{3}{4} .x = \dfrac{-1}{2}\)

=> \(x = \dfrac{-1}{2} : \dfrac{3}{4}\)

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }