The Wikibooks community is developing a policy on the use of generative AI. Please review the draft policy and provide feedback on its talk page.
将本部分中的方程配平方，使其变为圆的方程形式。
x 2 + 2 x + y 2 = 5 {\displaystyle x^{2}+2x+y^{2}=5\,}
x 2 + 2 x = 5 − y 2 {\displaystyle x^{2}+2x=5-y^{2}\,}
x 2 + 2 x + 1 = 6 − y 2 {\displaystyle x^{2}+2x+1=6-y^{2}\,}
我们在两边都加 1，以便将等式右边写成一个单一的幂。
( x + 1 ) 2 = 6 − y 2 {\displaystyle (x+1)^{2}=6-y^{2}\,}
( x + 1 ) 2 + y 2 = 6 {\displaystyle (x+1)^{2}+y^{2}=6\,}
x 2 + 4 x + y 2 − 4 y = 20 {\displaystyle x^{2}+4x+y^{2}-4y=20\,}
x 2 + 4 x = − ( y 2 − 4 y ) + 20 {\displaystyle x^{2}+4x=-(y^{2}-4y)+20\,}
x 2 + 4 x + 4 = − ( y 2 − 4 y ) + 20 {\displaystyle x^{2}+4x+4=-(y^{2}-4y)+20\,}
( x + 2 ) 2 = − ( y 2 − 4 y ) + 24 {\displaystyle (x+2)^{2}=-(y^{2}-4y)+24\,}
y 2 − 4 y = − ( x + 2 ) 2 + 24 {\displaystyle y^{2}-4y=-(x+2)^{2}+24\,}
y 2 − 4 y + 4 = − ( x + 2 ) 2 + 28 {\displaystyle y^{2}-4y+4=-(x+2)^{2}+28\,}
( y − 2 ) 2 + ( x + 2 ) 2 = 28 {\displaystyle (y-2)^{2}+(x+2)^{2}=28\,}
将不等式两边的分子和分母分别乘以不等式两边。
( 1 + t 2 ) ( 1 − s 2 ) > ( 1 − t 2 ) ( 1 + s 2 ) {\displaystyle (1+t^{2})(1-s^{2})>(1-t^{2})(1+s^{2})\,}
然后，展开并简化
1 − s 2 + t 2 − ( t s ) 2 > 1 + s 2 − t 2 − ( t s ) 2 {\displaystyle 1-s^{2}+t^{2}-(ts)^{2}>1+s^{2}-t^{2}-(ts)^{2}\,}
− s 2 + t 2 > s 2 − t 2 {\displaystyle -s^{2}+t^{2}>s^{2}-t^{2}\,}
− 2 s 2 + t 2 > − t 2 {\displaystyle -2s^{2}+t^{2}>-t^{2}\,}
− 2 s 2 > − 2 t 2 {\displaystyle -2s^{2}>-2t^{2}\,}
s 2 < t 2 {\displaystyle s^{2}<t^{2}\,}
s < t {\displaystyle s<t\,}
The Wikibooks community is developing a policy on the use of generative AI. Please review the draft policy and provide feedback on its talk page.