Câu 1:

\(\dfrac{-5}{6}:\dfrac{3}{13}=\dfrac{-5}{6}.\dfrac{13}{3}=\dfrac{-5.13}{6.3}=\dfrac{-65}{18}\) 

Câu 2:

\(\dfrac{5}{9}:\dfrac{5}{-3}=\dfrac{5}{9}.\dfrac{-3}{5}=\dfrac{5.-3}{9.5}=\dfrac{-15}{45}=\dfrac{-1}{3}\) 

Câu 3:

\(\left(-15\right):\dfrac{3}{2}=\left(-15\right).\dfrac{2}{3}=\dfrac{-15.2}{3}=\dfrac{-30}{3}=-10\) 

Câu 4:

\(\dfrac{3}{4}:\left(-9\right)=\dfrac{3}{4}.\dfrac{-1}{9}=\dfrac{3.-1}{4.9}=\dfrac{-3}{36}=\dfrac{-1}{12}\)

Câu 1

\(-\dfrac{5}{6}:\dfrac{3}{13}=-\dfrac{5}{6}.\dfrac{13}{3}=\dfrac{-5.13}{6.3}=-\dfrac{65}{18}\)

Câu 2

\(2:\dfrac{5}{9}:\dfrac{5}{-3}=2.\dfrac{9}{5}.\dfrac{-3}{5}=\dfrac{2.9.\left(-3\right)}{5.5}=-\dfrac{54}{25}\)

Câu 3

\(4:\dfrac{3}{2}:\left(-9\right)=4.\dfrac{2}{3}.\dfrac{1}{-9}=\dfrac{4.2.1}{-9}=-\dfrac{8}{9}\)

khỏi chép đề nha

a\Câu 1:

200 : ( 15 - x ) = 15 + 5

200 : ( 15 - x ) = 20

15 - x = 200 : 20

15 - x = 10

x = 15 - 10

x = 5

Câu 2:

63 : ( x - 5 ) = 22 -1

63 : ( x - 5 ) = 21

x - 5 = 63 : 21

x - 5 = 3

x = 3 + 5

x = 8

bái phục, bài quá nhiều!

Câu 13 :

\(\left(-\frac{1}{4}+\frac{5}{8}\right)+-\frac{3}{5}\)

\(=\frac{3}{8}-\frac{-3}{5}\)

\(=\frac{39}{40}\)

Câu 14 :

\(M=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)

\(=\frac{5}{9}.1=\frac{5}{9}\)

Câu 15 :

\(E=\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=0\)

Câu 16 :

\(H=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}+\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\frac{-7}{24}=-3\)

huhuhuhuhuhuhuhuhuhuhuhuhu giúp mk đi 

Plssssssssssssssssssssssssssssssssssssssssssssss

Câu 1: 

\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\) 

Câu 2:

\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\) 

Câu 3:

\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\) 

Câu 4:

\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\) 

\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\) 

\(=\dfrac{2}{7}.1\) 

\(=\dfrac{2}{7}\)

Câu 1

\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)

Câu 2

-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)

Câu 3

\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)

Câu 4

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

Em nên gõ công thức trực quan để đề bài rõ ràng nhé

Câu 1:

a. $3\frac{2}{3}+2\frac{1}{2}=(3+2)+(\frac{2}{3}+\frac{1}{2})=5+\frac{7}{6}=6+\frac{1}{6}=6\frac{1}{6}$

b. \(2\frac{1}{2}\times 3\frac{2}{5}=\frac{5}{2}\times \frac{17}{5}=\frac{17}{2}\)

c.

\(3\frac{1}{3}: 4\frac{1}{4}=\frac{10}{3}: \frac{17}{4}=\frac{40}{51}\)

d.

\(3\frac{1}{2}+4\frac{5}{7}-5\frac{5}{14}=(3+4-5)+(\frac{1}{2}+\frac{5}{7}-\frac{5}{14})=2+\frac{6}{7}=2\frac{6}{7}\)

Câu 2:

a. $x\times \frac{2}{7}=\frac{6}{11}$

$x=\frac{6}{11}: \frac{2}{7}=\frac{21}{11}$

b. $x: \frac{3}{2}=\frac{1}{4}$

$x=\frac{1}{4}\times \frac{3}{2}=\frac{3}{8}$

Câu 1 : x+1=1 => x = 0 => pt trên =-1 loại

              x+1 = 3 => x= 2 => 2y-1=3 => y=2

vậy x=2;y=2

câu 2 : 2x-1 = 1 > x = 1 ;  y +4=7 => y=3

           2x-1 = 7 => x=4 ; y +7 = 1 => y = -6 loại

vậy x=1, y=3 v

Câu 4  tương tự x=0 y=3

\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)

\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)

\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)

Thay vào BT

\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)