Bài 1:

$0,2-\frac{4}{7}+\frac{-6}{5}=\frac{1}{5}+\frac{-6}{5}-\frac{4}{7}$

$=\frac{-5}{5}-\frac{4}{7}=-1-\frac{4}{7}=\frac{-11}{7}$

b.

$=(\frac{-2}{3})^2+\frac{5}{6}+(-1)=\frac{4}{9}+\frac{5}{6}-1$

$=\frac{8}{18}+\frac{15}{18}-1=\frac{23}{18}-1=\frac{5}{18}$

c.

$=1+3+5+7+9=25$

Bài 2:

a. $-(0,5+x)-\frac{1}{3}=\frac{1}{6}$

$-(0,5+x)=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

$0,5+x=\frac{-1}{2}$

$x=\frac{-1}{2}-0,5=-1$
b.

$(x+\frac{4}{9})(x-\frac{11}{5})=0$

$\Rightarrow x+\frac{4}{9}=0$ hoặc $x-\frac{11}{5}=0$

$\Rightarrow x=\frac{-4}{9}$ hoặc $x=\frac{11}{5}$

c.

$\frac{1}{3}-|\frac{5}{4}-2x|=\frac{1}{4}$

$|\frac{5}{4}-2x|=\frac{1}{3}-\frac{1}{4}=\frac{1}{12}$
$\Rightarrow \frac{5}{4}-2x=\frac{1}{12}$ hoặc $\frac{5}{4}-2x=\frac{-1}{12}$

$\Rightarrow x=\frac{1}{2}(\frac{5}{4}-\frac{1}{12})$ hoặc $x=\frac{1}{2}(\frac{5}{4}+\frac{1}{12})$

$\Rightarrow x=\frac{7}{12}$ hoặc $x=\frac{2}{3}$

Câu 9

a) 2/15 + (-3/5)

= 2/15 - 3/5

= 2/15 - 9/15

= -7/15

b) -5/7 + 3/7 . (-2/9)

= -5/7 - 2/21

= -15/21 - 2/21

= -17/21

c) 5/16 - (7/15 - 3/16) + 17/30

= 5/16 - 7/15 + 3/16 + 17/30

= (5/16 + 3/16) + (17/30 - 14/30)

= 1/2 + 1/10

= 5/10 + 1/10

= 6/10

= 3/5

\(a,\dfrac{2}{15}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{15}-\dfrac{3}{5}\\ =\dfrac{2}{15}-\dfrac{9}{15}\\ =-\dfrac{7}{15}\\ b,-\dfrac{5}{7}+\dfrac{3}{7}\times\left(-\dfrac{2}{9}\right)\\ =-\dfrac{5}{7}-\dfrac{2}{21}\\ =-\dfrac{15}{21}-\dfrac{2}{21}\\ =-\dfrac{17}{21}\\ c,\dfrac{5}{16}-\left(\dfrac{7}{15}-\dfrac{3}{16}\right)+\dfrac{17}{30}\\ =\dfrac{5}{16}-\dfrac{7}{15}+\dfrac{3}{16}+\dfrac{17}{30}\\ =\left(\dfrac{5}{16}+\dfrac{3}{16}\right)-\left(\dfrac{17}{15}+\dfrac{17}{30}\right)\\ =\dfrac{1}{2}-\dfrac{17}{10}\\ =\dfrac{5}{10}-\dfrac{17}{10}\\ =-\dfrac{6}{5}.\)

\(\frac{6}{11}\cdot\frac{3}{14}+\frac{-5}{16}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-3}{16}\)

\(=\frac{6}{11}\cdot\frac{3}{14}+\frac{3}{14}\cdot\frac{5}{11}+\frac{-5}{16}+\frac{-3}{16}\)

\(=\frac{3}{14}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+\left(\frac{-5}{16}+\frac{-3}{16}\right)\)

\(=\frac{3}{14}\cdot\frac{6+5}{11}+\frac{-5+\left(-3\right)}{16}\)

\(=\frac{3}{14}\cdot\frac{11}{11}+\frac{-8}{16}\)

\(=\frac{3}{14}\cdot1+\frac{-1}{2}\)

\(=\frac{3}{14}+\frac{-1}{2}\)

\(=\frac{3}{14}+\frac{-7}{14}\)

\(=\frac{3+\left(-7\right)}{14}\)\(=\frac{-4}{14}=\frac{-2}{7}\)

\(\frac{-5}{6}+\left(7x-\frac{1}{2}\right)\cdot\frac{2}{9}=-1\frac{1}{3}\)

       \(\frac{-5}{6}+\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}\)

                       \(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{4}{3}+\frac{5}{6}\)

                       \(\frac{14}{9}\cdot x-\frac{1}{9}=-\frac{1}{2}\)

                                   \(\frac{14}{9}\cdot x=-\frac{1}{2}+\frac{1}{9}\)

                                   \(\frac{14}{9}\cdot x=-\frac{7}{18}\)

                                              \(x=-\frac{7}{18}:\frac{14}{9}\)

                                              \(x=-\frac{1}{4}\)

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)

\(a.\dfrac{3^{27}}{9^6.3^{16}}=\dfrac{3^{27}}{3^{12}.3^{16}}=\dfrac{3^{27}}{3^{28}}=\dfrac{1}{3}\)

\(\left(x-\dfrac{5}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow x-\dfrac{5}{2}=\pm\dfrac{3}{2}\)

\(TH1:x-\dfrac{5}{2}=\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\)

\(TH2:x-\dfrac{5}{2}=-\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\)

a: \(=\dfrac{3^{27}}{3^{12}\cdot3^{16}}=\dfrac{1}{3}\)

 câu b bài 2 thiếu nha e

B1

a, 638 +780 x 5 - 369 : 9 = 4497 

b, ( 273 + 485 ) x16 - 483 :3 x4 = 11484 

B2

a, 325 x 6 + 6 x 560 + 115 = 5425

B3

a, x = 1020

b, x = 36

B4

26 nha

ta có số cần tìm là xy , số mới là 5xy ( lưu ý là 5xy và xy là số )

5xy : xy = 26

ta có : 5x : 26 = 2 ( vì nếu ko sẽ ko ra k quả có số đầu là 5 ) 

 26 x 2 = 52

=> 52 - 52 hết và x = 2

y : 26 = 0 ( vì y < 26 )

=> ta có y= 0 

vậy số cần tìm là : xy = 20 

a: \(=\dfrac{216+3\cdot36+27}{13}=\dfrac{351}{13}=27\)

b: \(=\dfrac{1140-200-40}{9}+3^4=900+81=981\)

a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)

\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)

\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)

\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)

\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)

\(=\dfrac{13-x}{x}\)

c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)

\(=\dfrac{6\left(x+3\right)-2x+6-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)