Đáp án:

 Bài 1:

1, x(1-x)+$(x-1)^{2}$

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

=1-x

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-($x^{2}$ -3x+x-3)

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

=6x+7

4,(x+4)(x-2)-$x^{2}$ -$(x-3)^{2}$

=$x^{2}$ +4x-2x-8-($x^{2}$ -6x+9)

=$x^{2}$+4x-2x-8- $x^{2}$ +6x-9

=8x-17 

5,$(x-2)^{2}$ +(x-1)(x+5

)=$x^{2}$-4x+4+$x^{2}$-x+5x-5

=2$x^{2}$-1 

6,(x+3)(x-3)-x(23+x)

=$x^{2}$-9-(23x+$x^{2}$ )

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

=-23x-9

7, (1-2x)(5-3x)+$(4-x)^{2}$

=5-10x-3x+$6x^{2}$+16-8x+ $x^{2}$

=$7x^{2}$ -21x+21

8,(x-2)(x+2)-(x-3)(x+1)

=$x^{2}$-4-($x^{2}$ -3x+x-3)

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

=2x-1

Bài 2:

1,$x^{2}$ -9=0

(x-3)(x+3)=0

\(\left[ \begin{array}{l}x-3=0\\x+3=0\end{array} \right.\) 

\(\left[ \begin{array}{l}x=3\\x=-3\end{array} \right.\) 

2,   25-$x^{2}$ =0

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

\(\left[ \begin{array}{l}5-x=0\\5+x=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=5\\x=-5\end{array} \right.\) 

3,   -$x^{2}$ +36=0

(6-x)(6+x)=0

\(\left[ \begin{array}{l}6-x=0\\6+x=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=6\\x=-6\end{array} \right.\) 

4  $(3x+1)^{2}$ -16=0

(3x+1-4)(3x+1+4)=0

\(\left[ \begin{array}{l}3x-3=0\\3x+5=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=1\\x=-\frac{5}{3}\end{array} \right.\)

5,   $(2x-3)^{2}$ -49=0 

(2x-3-7)(2x-3+7)=0

\(\left[ \begin{array}{l}2x-10=0\\2x+4=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=5\\x=-2\end{array} \right.\)  

6,  $(2x-5)^{2}$ -$x^{2}$ =0

(2x-5-x)(2x-5+x)=0

\(\left[ \begin{array}{l}x-5=0\\3x-5=0\end{array} \right.\)

\(\left[ \begin{array}{l}x=5\\x=\frac{5}{3} \end{array} \right.\)