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

⇔ \(\dfrac{x-1}{3}=\dfrac{x-2}{2}\)

⇔ \(3x-6-2x+2=0\)

⇔ \(x-4=0\)

⇒ \(x=4\)

=>-2x+2=6-3x

=>-2x+3x=6-2

=>x=4

Buổi sáng mẹ bán được số quả trứng là :

   \(600\times\dfrac{2}{5}=240\left(quả\right)\)

Buổi chiều mẹ bán được số quả trứng là

  \(600\times\dfrac{1}{6}=100\left(quả\right)\)

Còn lại số quả trứng là

 \(600-240-100=260\left(quả\right)\)

    Đáp số 260 quả trứng

Buổi sáng bán được số quả trứng là:
`600 xx 2/5 =240(quả)`

Buổi chiều bán được số quả trứng là:

`600 xx 1/6 =100(quả)`

Còn lại số quả trứng là:

`600-240-100=260(quả)`

Đáp số: 260 qủa trứng

Buổi sáng bán là:

600x2/5=240(quả)

Buổi chiều bán là:

600x1/6=100(quả)

Còn lại số quả trứng là:

600-240-100=260(quả)

16/15

15

\(a,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{7}{5}-\dfrac{1}{3}=\dfrac{21}{15}-\dfrac{5}{15}=\dfrac{16}{15}\)

\(b,x:5=\dfrac{11}{6}\)

\(x=\dfrac{11}{6}\times5\)

\(x=\dfrac{55}{6}\)

7/5-1/2x2/3

=7/5-1/3

=21/15-5/15

=16/15

Phần diện tích làm bồn hoa chiếm:

1-2/7-3/5=4/35(tổng diện tích)

a) \(\dfrac{2}{3}+\dfrac{4}{15}=\dfrac{10}{15}+\dfrac{4}{15}=\dfrac{14}{15}\)

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

c) \(\dfrac{5}{3}\times\dfrac{9}{5}=\dfrac{45}{15}=3\)

d) \(\dfrac{3}{7}:2=\dfrac{3}{7}\times\dfrac{1}{2}=\dfrac{3}{14}\)

a  2/3 + 4/15= 10/15 + 4/15= 14/15

b, 9/10 - 2/5= 9/10 - 4/10 = 5/10 = 1/2

c, 5/3 x 9/5 = 9/3 = 3

d, 3/7 : 2= 3/7 : 2/1 = 3/7 x 1/2 = 3/14

Đg làm bài thi nên ko ai trl nha ^^

`7/12 + 3/4 xx 2/9`

`= 7/12 + 1/6`

`= 9/12`

_________________

`8/9 - 4/15 : 2/5`

`= 8/9 - 2/3`

`= 2/9`

giúp me 

a) _{^{\(\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{3}{4}\)}}

b) _{^{\(\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)}}

Lời giải:
a. $2^3.8=2^3.2^3=2^6$

b. $5^2.25=5^2.5^2=5^4$

c. $27:3^2=3^3:3^2=3^1$

d. $4^2.16=4^2.4^2=4^4$
e. $5^3.5^6=5^9$

f. $3^4.3=3^5$

g.$3^5.4^5=(3.4)^5=12^5$
h. $8^5.2^3=(2^3)^5.2^3=2^{15}.2^3=2^{18}$

i. $a^3.a^5=a^8$

j. $x^7.x.x^4=x^{7+1+4}=x^{12}$

k. $5^6:5^3=5^3$
l. $3^{15}:3^3=3^{15-3}=3^{12}$

m. $4^6:4^6=4^0=1$
n. $9^8:3^2=(3^2)^8:3^2=3^{16}:3^2=3^{14}$

o. $a:a=a^0=1$
p. $5^8.5.5^2=5^{8+1+2}=5^{11}$
q. $4.4^3=4^4$
r. $3.3^4=3^5$

s. $36.6^5=6^2.6^5=6^7$

t. $2^5.2^3=2^8$

u. $3^{10}:3^3=3^7$

v. $2^{10}:2^3=2^7$

w. $5^8:25=5^8:5^2=5^6$
x. $16:2^3=2^4:2^3=2^1$
y. $4.2^3=2^2.2^3=2^5$
z. $2^{10}:4=2^{10}:2^2=2^8$

\(A=\dfrac{31\cdot\left(31^{12}-1\right)}{31\left(31^{13}+1\right)}=\dfrac{31^{13}+1-32}{31\left(31^{13}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{14}+31}\)

\(B=\dfrac{31\left(31^{13}-1\right)}{31\left(31^{14}+1\right)}=\dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Dễ thấy \(31^{14}+31< 31^{15}+31\Rightarrow\dfrac{32}{31^{14}+31}>\dfrac{32}{31^{15}+31}\\ \Rightarrow\dfrac{1}{31}-\dfrac{32}{31^{14}+31}< \dfrac{1}{31}-\dfrac{32}{31^{15}+31}\)

Vậy A < B

\(I=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2009.2010}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2010}\\ =1-\dfrac{1}{2010}\\ =\dfrac{2009}{2010}\)

\(K=\dfrac{4}{2.4}+\dfrac{4}{4.6}+...+\dfrac{4}{2008.2010}\\ =2\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2008.2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\\ =2\left(\dfrac{1}{2}-\dfrac{1}{2010}\right)\\ =2.\dfrac{502}{1005}\\ =\dfrac{1004}{1005}\)

\(F=\dfrac{1}{18}+\dfrac{1}{54}+...+\dfrac{1}{990}\\ =\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{30.33}\\ =\dfrac{1}{3}\left(\dfrac{3}{3.6}+\dfrac{3}{6.9}+...+\dfrac{3}{30.33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{33}\right)\\ =\dfrac{1}{3}.\dfrac{10}{33}\\ =\dfrac{10}{99}\)

\(I=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2009}-\dfrac{1}{2010}=\dfrac{2010-1}{2010}=\dfrac{2008}{2010}\)

\(K=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}=\dfrac{502}{1005}\)

\(F=\dfrac{1}{3.6}+\dfrac{1}{6.9}+...+\dfrac{1}{99.100}=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{100}\right)=\dfrac{97}{900}\)