Lời giải:

$A=9+2.3^2+2.3^3+2.3^4+...+2.3^{2023}$

$A-9=2(3^2+3^3+3^4+...+3^{2023})$

$3(A-9)=2(3^3+3^4+3^5+...+3^{2024})$

$\Rightarrow 3(A-9)-(A-9)=2(3^{2024}-3^2)$

$2(A-9)=2.3^{2024}-18$

$\Rightarrow 2A-18=2.3^{2024}-18$

$\Rightarrow A=3^{2024}\vdots 3^{2023}$ (đpcm)

1/2x2/3+1/3x1/2

=1/2x(2/3+1/3)

=1/2x1

=1/2

2 x 5 + 5 x 7 + 9 x 3 

= 5 x ( 2 + 7 ) + 9 x 3 

= 5 x 9 + 9 x 3 

= 9 x ( 5 + 3 )

= 9 x 8

= 72

2x5+5x9+9x3                                       15:5+27:5+8:5

=9x5+(2+3)                                         =(15+8+27):5

=9x5+5                                               =50:5

=9x5+5x1                                            =10

=5x(9+1)                                             9+9x3+18:2x6

=5x10                                               =1x9+9x3+9x6

=50                                                   =9x(1+6+3)

                                                         =9x10

                                                         =90

\(15:5+27:5+8:5\)

\(=15\times\frac{1}{5}+27\times\frac{1}{5}+8\times\frac{1}{5}\)

\(=\left(15+27+8\right)\times\frac{1}{5}\)

\(=50\times\frac{1}{5}\)

\(=10\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\left(\frac{1}{1.2}-\frac{1}{9.10}\right).x=\frac{23}{45}\)

\(\frac{22}{45}.x=\frac{23}{45}\)

\(x=\frac{23}{45}:\frac{22}{45}\)

\(x=\frac{23}{22}\)

\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)

\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right).\frac{x}{2}=\frac{23}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{46}{45}\)

\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{46}{45}\)

\(\Rightarrow\frac{22}{45}.x=\frac{46}{45}\Rightarrow x=\frac{46}{22}=\frac{23}{11}\)

=>1/2x=-5/9+2/3=1/9

=>x=1/9*2=2/9

1: =-15+10-35+28=-12

3: =20-12-8+12=12

2) -3(4 - 7) + 5(-3 + 2)

= -3.4 + 3.7 - 5.3 + 5.2

= -12 + 21 -15 + 10

= 31 - 27

= 4

4) -5(2 - 7) + 4(2 - 5)

= -5.2 + 5.7 + 4.2 - 4.5

= -10 + 35 + 8 - 20

= 38 - 30

= 8

5) 6(-3 - 7) - 7(3 + 5)

= -6.3 - 6.7 - 7.3 - 7.5

= -18 - 42 - 21 - 35

= -116

6) 3(-5 + 6) - 4(3 - 2)

= -3.5 + 3.6 - 4.3 + 4.2

= -15 + 18 - 12 + 8

= 26 - 27

= -1

Bài 1 :

Để \(x^2+5x-x^2\)

\(\Leftrightarrow5>-x^2+x\)

À quên mọi người tính nhanh nhé !

a) \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}\)

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

\(=\frac{3}{4}\)

b) \(\frac{4}{7}\times\frac{8}{9}+\frac{4}{7}\times\frac{1}{9}\)

\(=\frac{4}{7}\times\left(\frac{8}{9}+\frac{1}{9}\right)\)

\(=\frac{4}{7}\times1\)

\(=\frac{4}{7}\)

c) \(\frac{13}{6}\times\frac{3}{8}-\frac{3}{8}\times\frac{7}{6}\)

\(=\frac{3}{8}\times\left(\frac{13}{6}-\frac{7}{6}\right)\)

\(=\frac{3}{8}\times1\)

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

d) \(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{9x10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

a) Bạn tự làm :V

b) \(\frac{4}{7}\times\frac{8}{9}+\frac{4}{7}\times\frac{1}{9}\)

\(=\frac{4}{7}\times\left(\frac{8}{9}+\frac{1}{9}\right)\)

\(=\frac{4}{7}\times1\)

\(=\frac{4}{7}\)

c) \(\frac{13}{6}\times\frac{3}{8}-\frac{3}{8}\times\frac{7}{6}\)

\(=\frac{3}{8}\times\left(\frac{13}{6}-\frac{7}{6}\right)\)

\(=\frac{3}{8}\times1\)

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

d) Bạn tự làm :v