bạn viết rõ lũy thừa giúp mình với

\(A=B\)

\(2.\left(70-x\right)+2^3.3^2=92\)

\(140-2x+72=92\)

\(-2x=92-212\)

\(-2x=-120\)

\(x=60\)

\(\dfrac{-19}{23}\cdot\dfrac{13}{14}+\dfrac{13}{14}\cdot\dfrac{-15}{23}-\dfrac{13}{14}\cdot\dfrac{1}{23}\\ =\dfrac{13}{14}\cdot\left(\dfrac{-19}{23}+\dfrac{-15}{23}-\dfrac{1}{23}\right)\\ =\dfrac{13}{14}\cdot\dfrac{-35}{23}=\dfrac{-65}{46}\)

Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$

$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$

$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương 

$\Rightarrow x+2023=0$

$\Leftrightarrow x=-2023$

=35.14-35.23-23.14+23.35

=(-35.23+23.35)+(35.14-23.14)

=0+14.12

=168

Kết quả của bài này là 168. Có sai thì đừng...mắng mik nha!^_^

\(A=35-\frac{167}{32}\)

\(A=\frac{953}{32}\)

1) (-37) + 14 + 26 + 37

= [(-37) + 37] + 14 + 26

= 0 + 40

= 40

2) (-24) + 6 + 10 + 24

= [(-24) + 24] + 6 + 10

= 0 + 16

= 16

3) 15 + 23 + (-25) + (-23)

= 15 + (-25) + [(-23) + 23]

= -10 + 0

= -10

4) 60 + 33 + (-50) + (-33)

= 60 + (-50) + [(-33) + 33]

= 10 + 0

= 10

a) Ta có : B = 23! + 19! + 15! 

= 1.2.3.4..10.11...23 + 1.2.3.4...10.11..19 + 1.2.3.4..10.11..15

= 11(1.2.3...10.12.23 + 1.2.3.4..10.12...19 + 1.2.3.4....10.12...15) \(⋮\)11

b) Lại  có B =  23! + 19! + 15! 

= 1.2.3.4..10.11...23 + 1.2.3.4...10.11..19 + 1.2.3.4..10.11..15

= 10.11(1.2.3.4..9.12...23  + 1.2.3.4...9.12...19 + 1.2.3.4...9.12...15)

= 110(1.2.3.4..9.12...23  + 1.2.3.4...9.12...19 + 1.2.3.4...9.12...15) \(⋮\)110

               BÀI GIẢI:

a) Ta có: B = 23! + 19! + 15!

= 1.2.3. ... .10 .11. ... 23 + 1.2.3. ... .10.11. ... .19 + 1.2.3. ... .10.11. ... .15

= 11 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15) \(⋮11\)

b) Ta có: B = 23! + 19! + 15!

= 1.2.3. ... .10 .11. ... 23 + 1.2.3. ... .10.11. ... .19 + 1.2.3. ... .10.11. ... .15

= 10.11 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15)

= 110 (1.2.3. ... .10.12. ... .23 + 1.2.3. ... .10.12. ... .19 + 1.2.3. ... .10.12. ... .15) \(⋮110\)