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

a, Ta có : \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|=10x-100\)

Ta thấy : \(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\\left|...\right|\ge0\\\left|x-9\right|\ge0\end{matrix}\right.\)

=> \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|\ge0\)

=> \(10x-100\ge0\)

=> \(x\ge10\)

=> \(\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|...\right|=...\end{matrix}\right.\)

=> \(x-1+x-2+...+x-9=10x-100\)

=> \(9x-45=10x-100\)

=> \(10x-9x=100-45\)

=> \(x=55\) ( TM )

Vậy ....

b, Ta có : \(\left|x-2\right|+...+\left|x-9\right|=1-x\)

Ta thấy : \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|...\right|\ge0\end{matrix}\right.\)

=> \(\left|x-2\right|+...+\left|x-9\right|\ge0\)

=> \(1-x\ge0\)

=> \(x\le1\)

=> \(\left\{{}\begin{matrix}\left|x-2\right|=2-x\\\left|...\right|=-...\end{matrix}\right.\)

=> \(2-x+...+9-x=1-x\)

=> \(44-8x=1-x\)

=> \(8x-x=44-1\)

=> \(x=\frac{43}{7}\) ( KTM )

Vậy phương trình vô nghiệm .

`@` `\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, \(25\left(x-2\right)^2-100\left(y-3\right)^2\)

\(=\left(5x-10\right)^2-\left(10y-30\right)^2\)

\(=\left(5x-10-10y+30\right)\left(5x-10+10y-30\right)\)

\(=\left(5x-10y+20\right)\left(5x+10y-40\right)\)

\(=25\left(x-2y+4\right)\left(x+2y-8\right)\)

b, \(4\left(x+3\right)^2-9\left(x+2\right)^2\)

\(=\left(2x+6\right)^2-\left(3x+6\right)^2\)

\(=\left(2x+6-3x-6\right)\left(2x+6+3x+6\right)\)

\(=-x\left(5x+12\right)\)

c, \(x^2-10x-y^2+25\)

\(=\left(x-5\right)^2-y^2\)

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

d, \(x^2-4x-y^2+4\)

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

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

e, \(\left(x^2+2\right)^2-2\left(x^2+2\right)+1\)

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

\(=\left(x^2+1\right)^2\)

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

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

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

\(D=0+1\)

\(D=1\)