10. (1)(2023·贵州)先化简,再求值：$(1-x+\frac{2x-1}{x+1})÷\frac{x-2}{x^{2}+2x+1}$,其中$x的值是一元二次方程x^{2}-2x-3= 0$的解.
(2)(2023·达州)已知关于$x的一元二次方程kx^{2}+(2k+1)x+2= 0$的解为整数,求整数$k$取值的个数.

(1)【解】$(1-x+\frac {2x-1}{x+1})÷\frac {x-2}{x^{2}+2x+1}$
$=\frac {(1-x)(1+x)+2x-1}{x+1}÷\frac {x-2}{(x+1)^{2}}$
$=\frac {1-x^{2}+2x-1}{x+1}\cdot \frac {(x+1)^{2}}{x-2}$
$=\frac {-x^{2}+2x}{x+1}\cdot \frac {(x+1)^{2}}{x-2}$
$=\frac {-x(x-2)}{x+1}\cdot \frac {(x+1)^{2}}{x-2}$
$=-x(x+1)$
$=-x^{2}-x.$
$\because x^{2}-2x-3=0,\therefore x_{1}=3,x_{2}=-1.$
$\because x+1≠0,x-2≠0,\therefore x=3,$
$\therefore$ 原式$=-x^{2}-x=-9-3=-12.$
(2)【解】$\because kx^{2}+(2k+1)x+2=0,$
$\therefore (kx+1)(x+2)=0,$
解得$x_{1}=-\frac {1}{k},x_{2}=-2.$
$\because$ 关于$x$的一元二次方程$kx^{2}+(2k+1)x+2=0$的解为整数,$\therefore k=\pm 1,$
故整数$k$取值的个数为2.

11. (1)(2025·编写)若直角三角形的两边长分别是方程$x^{2}-7x+12= 0$的两根,则该直角三角形的面积是______.
(2)(2025·编写)若方程$mx^{2}+nx-6= 0的两根为x_{1}= 3,x_{2}= -2$,则方程$m(y-1)^{2}+n|y-1|-6= 0$的根为______.

(1)6或$\frac {3\sqrt {7}}{2}$ (2)$y_{1}=4,y_{2}=-2$

12. (1)(2025·编写)若$(m^{2}-n^{2})(m^{2}-n^{2}-2)-8= 0$,则$m^{2}-n^{2}$的值是______.
(2)(2025·编写)若$(x^{2}+y^{2})^{2}-3(x^{2}+y^{2})-70= 0$,则$x^{2}+y^{2}= $______.

(1)4或-2 (2)10