\(M=x^{10}-25x^9+25x^8-25x^7+...-25x^3+25x^2-25x+25\)

Ta thấy : \(x=24\Rightarrow x+1=25\)

\(\Rightarrow M=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

\(M=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^4-x^3+x^3+x^2-x^2-x+x+1\)

\(\Rightarrow M=1\)

Vậy \(M=1\left(tạix=24\right)\)

M=x 

10

 −25x 

9

 +25x 

8

 −25x 

7

 +...−25x 

3

 +25x 

2

 −25x+25

Ta thấy : 

x

=

24

⇒

x

+

1

=

25

x=24⇒x+1=25

⇒

M

=

x

10

−

(

x

+

1

)

x

9

+

(

x

+

1

)

x

8

−

(

x

+

1

)

x

7

+

.

.

.

−

(

x

+

1

)

x

3

+

(

x

+

1

)

x

2

−

(

x

+

1

)

x

+

(

x

+

1

)

⇒M=x 

10

 −(x+1)x 

9

 +(x+1)x 

8

 −(x+1)x 

7

 +...−(x+1)x 

3

 +(x+1)x 

2

 −(x+1)x+(x+1)

M

=

x

10

−

x

10

−

x

9

+

x

9

+

x

8

−

x

8

−

x

7

+

.

.

.

−

x

4

−

x

3

+

x

3

+

x

2

−

x

2

−

x

+

x

+

1

M=x 

10

 −x 

10

 −x 

9

 +x 

9

 +x 

8

 −x 

8

 −x 

7

 +...−x 

4

 −x 

3

 +x 

3

 +x 

2

 −x 

2

 −x+x+1

⇒

M

=

1

⇒M=1

Vậy 

M

=

1

(

t

ạ

i

x

=

24

)

M=1(tạix=24)

\(M=x^{2023}-2023.\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

Ta có : \(x=2022\Rightarrow x+1=2023\)

\(\Rightarrow M=x^{2023}-\left(x+1\right).\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)

\(\Rightarrow M=x^{2023}-\left(x+1\right)x^{2022}+\left(x+1\right)x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}+...-\left(x+1\right)x^2+\left(x+1\right)x\)

\(\Rightarrow M=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}-x^{2021}-x^{2020}+x^{2020}+x^{2019}-x^{2019}-...-x^3-x^2+x^2+x\)

\(\Rightarrow M=x\)

\(\Rightarrow M=2022\)

Vậy \(M=2022\left(tạix=2022\right)\)

\(\left(x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-4\right)=0\)

\(\left(x-4\right)\left(x+3+x+5\right)=0\)

\(\left(x-4\right)\left(2x+8\right)=0\)

\(\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\\2x+8=0\Leftrightarrow x=-4\end{matrix}\right.\)

\(102-\left[8^2-48\right].0,5.\left(2^2.10+8\right):2^5\)

\(=102-\left[64-48\right].0,5.\left(40+8\right):32\)

\(=102-16.0,5.48:32\)

\(=102-\left(8.\dfrac{3}{2}\right)\)

\(=102-12\)

\(=90\)

cứu mình với

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)

\(=\dfrac{1}{3}-\dfrac{1}{99}\)

\(=\dfrac{33}{99}-\dfrac{1}{99}\)

\(=\dfrac{32}{99}\)

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{98.99}\\ =\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\\ =\dfrac{1}{3}-\dfrac{1}{99}\\ =\dfrac{32}{99}\)

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