bài1  

a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\) 

=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\) 

=\(\dfrac{1}{12}+\dfrac{9}{12}\) 

=\(\dfrac{10}{12}=\dfrac{5}{6}\)

bài 1 

b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\) 

= \(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\) 

= \(-\dfrac{6}{5}+\dfrac{3}{10}\) 

=\(-\dfrac{12}{10}+\dfrac{3}{10}\) 

=\(-\dfrac{9}{10}\) 

b: \(A=\dfrac{10^7-8+13}{10^7-8}=1+\dfrac{13}{10^7-8}\)

\(B=\dfrac{10^8-7+13}{10^8-7}=1+\dfrac{13}{10^8-7}\)

mà \(10^7-8< 10^8-7\)

nên A>B

c: \(\dfrac{1}{10}A=\dfrac{10^{1992}+1}{10^{1992}+10}=1-\dfrac{9}{10^{1992}+10}\)

\(\dfrac{1}{10}B=\dfrac{10^{1993}+1}{10^{1993}+10}=1-\dfrac{9}{10^{1993}+10}\)

mà \(\dfrac{9}{10^{1992}+10}>\dfrac{9}{10^{1993}+10}\)

nên A<B

a,A<B

b,A,<B

c,A<B

a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)

Vậy A < B

b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)

c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:

 \(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

Vậy A < B

a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\)\(x\)            = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\) \(x\)           = - \(\dfrac{3}{20}\)

                  \(x\)          =  - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)

                  \(x\)          = - \(\dfrac{3}{8}\)

b, \(\dfrac{1}{3}\) +   \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{1}{5}\)

            \(\dfrac{1}{2}\): \(x\) =  - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)

             \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)

                   \(x\) =  \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))

                    \(x\) =  - \(\dfrac{15}{16}\)

c, - \(\dfrac{2}{3}\): \(x\) + \(\dfrac{5}{8}\) = - \(\dfrac{7}{12}\)

     \(\dfrac{2}{3}\): \(x\)        =  \(\dfrac{7}{12}\) + \(\dfrac{5}{8}\)

     \(\dfrac{2}{3}\) : \(x\)     =  \(\dfrac{29}{24}\)

           \(x\)      = \(\dfrac{2}{3}\) : \(\dfrac{29}{24}\)

            \(x\)     = \(\dfrac{16}{29}\)

\(2\dfrac{1}{3}.3=\dfrac{7}{3}.3=7.\\ \left(\dfrac{2}{5}-\dfrac{3}{4}\right)-\dfrac{2}{5}=\dfrac{2}{5}-\dfrac{3}{4}-\dfrac{2}{5}=-\dfrac{3}{4}.\\ \dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}.\\ =\dfrac{-10}{11}\left(\dfrac{4}{7}+\dfrac{3}{7}-1\right).\\ =\dfrac{-10}{11}.\left(1-1\right)=0.\)

1) 2\(\dfrac{1}{3}\).3=\(\dfrac{7}{3}\).3=7.

2) (2/5 -3/4) -2/5 = 2/5 -3/4 -2/5 = -3/4.

3) \(\dfrac{-10}{11}.\dfrac{4}{7}+\dfrac{-10}{11}.\dfrac{3}{7}+1\dfrac{10}{11}=\dfrac{1}{11}\left(-\dfrac{40}{7}-\dfrac{30}{7}+21\right)=\dfrac{1}{11}.\left(-10+21\right)=1\).

a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)

=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)

=\(\dfrac{10}{11}.\dfrac{-1}{2}\)

=\(\dfrac{-5}{11}\)

b; 

B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8

B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8

B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8

B = \(\dfrac{2}{7}\) - 8

B = \(\dfrac{2}{7}-\dfrac{56}{7}\)

B = - \(\dfrac{54}{7}\)

c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))

C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))

C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]

C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\)  - \(\dfrac{588}{105}\)]

C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]

C = - 15.(- \(\dfrac{143}{105}\))

C = \(\dfrac{143}{7}\)

đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)

\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)

=2căn 5-2-2căn 5

=-2

d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)

\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)

`1/8 -1/2 + (-11/12 + 1)`

`=1/8 -1/2 + (-11/12 +12/12)`

`=1/8 -1/2 + 1/12`

`= 1/8- 4/8+1/12`

`= -3/8 + 1/12`

`=-7/24`

`---------`

`3/5 -(-7/10) + (-13/10)`

`= 3/5 + 7/10 + (-13/10)`

`= 6/10 + 7/10 + (-13/10)`

`= 13/10 +(-13/10)`

`= 0/10=0`

\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)\\ =\dfrac{-3}{8}+\dfrac{1}{12}\\ =\dfrac{-7}{24}\\ \dfrac{3}{5}-\dfrac{-7}{10}+\left(-\dfrac{13}{10}\right)\\ =\dfrac{13}{10}-\dfrac{13}{10}\\ =0\)

\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)=\dfrac{-3}{8}+\dfrac{1}{12}=\dfrac{-7}{24}\)

\(\dfrac{3}{5}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{6}{10}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{13}{10}+\dfrac{-13}{10}=\dfrac{0}{10}=0\)

=9/10-2/10+1

=7/10+10/10

=17/10

17/10

= 9/10 - 1/5 + 1

= 7/10 + 1

= 17/10

\(E=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{5751}{25}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)

\(=\dfrac{\dfrac{25}{108}.\dfrac{5751}{25}+\dfrac{187}{4}}{\dfrac{100}{21}:\left(\dfrac{-44}{21}\right)}\)

\(=\dfrac{53,25+\dfrac{187}{4}}{\dfrac{-25}{11}}\)

\(=\dfrac{100}{\dfrac{-25}{11}}\)

\(=-44\)