\(\sqrt{36x^4\left(b-2\right)}^2=6x^2\left(2-b\right)=12x^2-6x^2b\)

\(\sqrt{36x^4\left(b-2\right)^2}=6x^2\left|b-2\right|=6x^2\left(2-b\right)\) (vì b<2 nên b-2<0)

a: Ta có: \(25x^2\left(x-y\right)-x+y\)

\(=\left(x-y\right)\left(25x^2-1\right)\)

\(=\left(x-y\right)\left(5x-1\right)\left(5x+1\right)\)

b: Ta có: \(16x^2\left(z^2-y^2\right)-z^2+y^2\)

\(=\left(z^2-y^2\right)\left(16x^2-1\right)\)

\(=\left(z-y\right)\left(z+y\right)\left(4x-1\right)\left(4x+1\right)\)

c: Ta có: \(x^3+x^2y-x^2z-xyz\)

\(=x^2\left(x+y\right)-xz\left(x+y\right)\)

\(=x\left(x+y\right)\left(x-z\right)\)

Lời giải:

$M=(x^{10}-24x^9)-(x^9-24x^8)+(x^8-24x^7)-(x^7-24x^6)+(x^6-24x^5)-(x^5-24x^4)+(x^4-24x^3)-(x^3-24x^2)+(x^2-24x)-(x-24)+1$

$=x^9(x-24)-x^8(x-24)+x^7(x-24)-.....+x(x-24)-(x-24)+1$

$=(x-24)(x^9-x^8+x^7-...+x-1)+1$

$=0.(x^9-x^8+....+x-1)+1=1$

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

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)

\(2-25x^2=0\)

\(\Rightarrow25x^2=2\)

\(\Rightarrow x^2=\frac{2}{25}\)

\(\Rightarrow x=\frac{\sqrt{2}}{5}\)

tíc mình nha

\(2-25x^2=0\)

\(\Leftrightarrow\left(\sqrt{2}-5x\right)\left(\sqrt{2}+5x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2}-5x=0\\\sqrt{2}+5x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)

Vậy: \(x=\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{cases}}\)

b) \(x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

2x³y³z

2x³y³z

\(-\dfrac{2}{3}xy^2z.9x^2y^2=-6x^3y^4z\)

6xy4z

6xy4z

Biểu thức  x 2 - 25 x 2 - 10 x + 25 x xác định khi x 0 và x  ≠  5.

Ta có: khi x(x + 5) = 0 và x – 5  ≠  0

x(x + 5) = 0 ⇔ x = 0 hoặc x + 5 = 0 ⇔ x = - 5

x = 0 không thỏa mãn điều kiện.

Vậy x = - 5 thì biểu thức  x 2 - 25 x 2 - 10 x + 25 x có giá trị bằng 0.