x=24

=> x+1=25

=> M=x¹⁰⁰-(x+1)x⁹⁹+(x+1)x⁹⁸-(x+1)x⁹⁷+(x+1)x⁹⁶-...-(x+1)x³+(x+1)x²-(x+1)x+25

       =x¹⁰⁰-x¹⁰⁰-x⁹⁹+x⁹⁹+x⁹⁸-x⁹⁸-x⁹⁷+x⁹⁷+x⁹⁶-...-x⁴-x³+x³+x²-x²-x+25

       =-x+25

       =-24+25

       =1

Vậy M=1.

x=-24

=>-x=24

=>-x+1=25

thay -x+1=25 vào E ta được:

E=x²⁰+(-x+1)x¹⁹+(-x+1)x¹⁸+(-x+1)x¹⁷+...+(-x+1)x³+(-x+1)x²+(-x+1)x+(-x+1)

=x²⁰-x²⁰+x¹⁹-x¹⁹+x¹⁸-x¹⁸+x¹⁷-...-x⁴+x³-x³+x²-x²+x-x+1

=1

Vậy với x=-24 thì E=1

x = ‐24

=> ‐ X = 24

=> ‐ X + 1 = 25

thay ‐x+1=25 vào E ta được:

E = x 20 + ﴾‐ x + 1﴿ x 19 + ﴾‐ x + 1﴿ x 18 + ﴾‐ x + 1﴿ x 17 + ... + ﴾‐ x + 1﴿ x 3 + ﴾‐ x + 1 ﴿ x 2 + ﴾‐ x + 1﴿ x + ﴾‐ x + 1﴿

= x 20 ‐x 20 + x 19 ‐x 19 + x 1 8 ‐x 18 + x 17 ‐...‐ x 4 + x 3 ‐x 3 + x 2 ‐x 2 + x‐x + 1

= 1

Vậy với x=‐24 thì E=1

Học tốt nha Nguyễn Quang Linh

cảm ơn bạn...nhưng câu hỏi đó mik gửi từ ngày 22/7/2015 đến giờ sao bạn thấy mà trả lời vậy???

x=24 nên x+1=25

C=x^4-x^3(x+1)+x^2(x+1)-x(x+1)+30

=x^4-x^4-x^3+x^3+x^2-x^2-x+30

=-x+30

=30-24

=6

a, A = x⁵ - 5x⁴ + 5x³ - 5x² + 5x - 1

A= x⁵ - ( 4+1 ) x⁴ + ( 4+1 ) x³ - ( 4+1) x² + ( 4+1 ) x -1

Thay 4 = x vào biểu thức A, ta đc :

A = x⁵ - ( x+1 ) x⁴ + ( x+1 ) x³ - ( x+1 ) x² + ( x+1 ) x - 1

A = x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x -1

A = x -1

Thay x = 4 vào biểu thức A, ta đc :

A = 4 -1 

A = 3

b, B = x⁷ - 80x⁶ + 80x⁵ - 80x⁴ + .....+ 80x + 15

B = x⁷ - ( 79 +1 ) x⁶ + ( 79+1 )x⁵ - ( 79+1 ) x⁴ +....+( 79+1 )x + 15

Thay 79 = z vào biểu thức A, ta có :

B = x⁷ - ( x + 1 )x⁶ + ( x+1 )x⁵ - ( x+1 )x⁴ + .....+ ( x+1 )x +15

B= x⁷ - x⁷ - x⁶ + x⁶ + x⁵ - x⁵ - x⁴ + .....- x² + x² + x + 15

B= x + 15

Thay x= 79 vào biểu thức A, ta có:

A = 79 + 15

A= 94

c, C = x¹⁴ - 10x¹³ + 10x¹² - 10x¹¹ + ....+ 10x² - 10x + 10

C= x¹⁴ - ( x +1 )x¹³ + ( x + 1 ) x¹² - ( x + 1 )x¹¹ + ..... + ( x + 1 )x² - ( x + 1 )x - 10

C= x¹⁴ - x¹⁴ - x¹³ + x¹³ + x¹² - x¹² - x¹¹ +....+ x³ - x² + x² - x +10

C= -x -10 

Thay -x = -9 vào biểu thức C, ta có :

C = -9 + 10

C = 1

d, D = x¹⁰ - ( x+1 )x⁹ + (x + 1 )x⁸ - ( x+1 )x⁷ +....+( x+1 )x² - ( x + 1 )x + 25

D = x¹⁰ - ( x + 1 ) x⁹ + ( x + 1 )x⁸ - ( x + 1 )x⁷ + ..... + x³ - x² + x² - x + 25

D = -x + 25

thay -x = -24, vào biểu thức A , ta đc ;

A = -24 + 25

A = 1

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)

\(=x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x-x+3\)

\(=3\)

Ta có : 

\(A=x^5-5x^4+5x^3-5x^2+5x-1\)

\(A=x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x-x+3\)\(A=x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+3\)

\(A=3\)

P/s tham khảo nha 

hok tốt

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x-28x+3x+6=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-13x+x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\cdot\left(x+3\right)\left(6x^2-5x+1\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)

hay \(x\in\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)

Ta có: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3-3x^2+16x^2-8x-6x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[3x^2\left(2x-1\right)+8x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+9x-x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left[3x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(x+3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-1=0\\x+3=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=1\\x=-3\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\\x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{1}{2};-3;\dfrac{1}{3}\right\}\)

\(B=\sqrt{\left(5x-3\right)^2}+\sqrt{\left(5x-4\right)^2}\ge\left|5x-3\right|+\left|4-5x\right|\ge5x-3+4-5x=1\).

Dấu "=" xảy ra khi và chỉ khi \(3\le5x\le4\Leftrightarrow\dfrac{3}{5}\le x\le\dfrac{4}{5}\)