Bài làm:

Ta có: \(x=-9\Leftrightarrow-10=x-1\Rightarrow10=1-x\)nên thay vào ta tính:

\(P\left(-9\right)=1+\left(1-x\right)x+\left(1-x\right)x^2+\left(1-x\right)x^3+...+\left(1-x\right)x^{19}+\left(1-x\right)x^{20}\)

\(P\left(-9\right)=1+x-x^2+x^2-x^3+x^3-x^4+...+x^{20}-x^{21}\)

\(P\left(-9\right)=1+x-x^{21}\)

\(P\left(-9\right)=1-9+9^{21}\)

\(P\left(-9\right)=9^{21}-8\)

Vậy khi \(x=-9\)thì \(P\left(x\right)=9^{21}-8\)

Học tốt!!!!

a) (x+9)(x-9)-x²=x²-81-x²=-81

b) (10x-1)(10x+1)-(10x-1)²=100x²-1-100x²+20x-1=20x-2

d) (x-1)(x-2)-(x-2)(x+2)=x²-3x+2-x²+4=-3x+6

tks mn

a, 15x - 6 = 12x + 3

\(\Leftrightarrow\) 15x - 12x = 3 + 6

\(\Leftrightarrow\) 3x = 9

\(\Leftrightarrow\) x = 3

Vậy S = {3}

b, \(\frac{x+2}{2}-\frac{2x-3}{5}=10x+\frac{13}{10}\)

\(\Leftrightarrow\) \(\frac{5\left(x+2\right)}{10}-\frac{2\left(2x-3\right)}{10}=\frac{100x}{10}+\frac{13}{10}\)

\(\Leftrightarrow\) 5(x + 2) - 2(2x - 3) - 100x - 13 = 0

\(\Leftrightarrow\) 5x + 10 - 4x + 6 - 100x - 13 = 0

\(\Leftrightarrow\) -99x + 3 = 0

\(\Leftrightarrow\) x = \(\frac{1}{33}\)

Vậy S = {\(\frac{1}{33}\)}

d, (3x + 2)(4x - 5) = 0

\(\Leftrightarrow\) 3x + 2 = 0 hoặc 4x - 5 = 0

\(\Leftrightarrow\) x = \(\frac{-2}{3}\) và x = \(\frac{5}{4}\)

Vậy S = {\(\frac{-2}{3}\); \(\frac{5}{4}\)}

Phần c với phần e bạn viết vậy mình ko hiểu, bn viết lại đi!

Chúc bn học tốt!!

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a,x²-25-(x+5)                                                   b,mình quên mất rồi.Đợi tí nhé

(x²-25)-(x+5)=0

(x²-5²)+(x-5)=0

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

(x-5)(x+5+1)=0

x-5=0 hoặc x+5+1=0

x=0+5 hoặc x=0-5-1

x=5     hoặc x=-6

Vậy x=5 và x=-6

Giải bpt

A)  (x^2+1)×(4x-2)≫0(lớn hơn hoặc =0)

B) (x-2)×x^2>0

Mog mn giúp ạ

E cần gấp

Thak mn

x=9

=>x+1=10

\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)

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

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

=-x+1

=-9+1=-8

c) Đặt \(f\left(x\right)=x^{10}-10x+9\)

Giả sử \(f\left(x\right)⋮\left(x-1\right)^2\)

\(\Rightarrow f\left(x\right)=\left(x-1\right)^2Q\left(x\right)\)

\(\Leftrightarrow f\left(1\right)=\left(1-1\right)^2Q\left(1\right)\)

                  \(=0\)

\(\Leftrightarrow1^{10}-10.1+9=0\)

\(\Leftrightarrow0=0\)( đúng)

\(\Rightarrow\)điều giả sử đúng

\(\Rightarrow f\left(x\right)⋮\left(x-1\right)^2\left(đpcm\right)\)

e) \(E=x^5-15x^4+16x^3-29x^2+13x\) tại x = 14

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

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

\(E=-x\)

\(E=-14\)

d)  \(D=x^3-30x^2-31+1\) tại x = 31

\(D=31^3-30.31^2-31+1\)

\(D=31^2\left(31-30-1\right)+1\)

\(D=0+1\)

\(D=1\)