Giúp mình với! $J)~^{1-}(1+4y)$ $8):~-(-2x+3)^2-(5x-3)^2$ $9)~(-2x+3)^2-(5x-3$,Bài 3. Thực hiện phép tính,$1)~x(1-x)+(x-1)^2$ $2)~(x-3)^2-x^2+10x-7$,$3)~(x+2)^2-(x-3)(x+1)$ $4)~(x+4)(x-2)-(x-3)^2$,$5)~(x-2)^2+(x-1)(x+5)$ $6)~(x+3)(x-3)-x(23+x)$

Question

Ninh Hoàng · Accepted Answer

{"userId":5421384,"userName":"midnight"}

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

$$=x-x^2+x^2-2x+1$$

$$=-x+1$$

$$2)\left(x-3\right)^{2}-x^2+10x-7$$

$$=x^2-6x+9-x^2+10x-7$$

$$=4x+2$$

$$3)\left(x+2\right)^2-\left(x-3\right)\left(x+1\right)$$

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

$$=6x+7$$

$$4)\left(x+4\right)\left(x-2\right)-\left(x-3\right)^2$$

$$=x^2+2x-8-x^2+6x-9$$

$$=8x-17$$

$$5)\left(x-2\right)^2+\left(x-1\right)\left(x+5\right)$$

$$=x^2-4x+4+x^2+4x-5$$

$$=2x^2-1$$

$$6)\left(x+3\right)\left(x-3\right)-x\left(23+x\right)$$

$$=x^2-9-23x-x^2$$

$$=-23x-9$$.

Timi · Answer

Bài 3:
1) $ x(1-x) + (x-1)^2 $
   $ = x - x^2 + x^2 - 2x + 1 $
   $ = -x + 1 $

2) $ (x-3)^2 - x^2 + 10x - 7 $
   $ = x^2 - 6x + 9 - x^2 + 10x - 7 $
   $ = 4x + 2 $

3) $ (x+2)^2 - (x-3)(x+1) $
   $ = x^2 + 4x + 4 - (x^2 - 2x - 3) $
   $ = x^2 + 4x + 4 - x^2 + 2x + 3 $
   $ = 6x + 7 $

4) $ (x+4)(x-2) - (x-3)^2 $
   $ = x^2 + 2x - 8 - (x^2 - 6x + 9) $
   $ = x^2 + 2x - 8 - x^2 + 6x - 9 $
   $ = 8x - 17 $

5) $ (x-2)^2 + (x-1)(x+5) $
   $ = x^2 - 4x + 4 + x^2 + 4x - 5 $
   $ = 2x^2 - 1 $

6) $ (x+3)(x-3) - x(23+x) $
   $ = x^2 - 9 - 23x - x^2 $
   $ = -23x - 9 $