初三数学,求学霸。 50
1个回答
展开全部
1) 韦达定理:
alpha + beta = - 5/2
alpha . beta = 1 / 2
==》
alpha^2 + alpha beta + beta^2
= (alpha + beta)^2 - alpha beta = 25/4 - 1/2 = 23 / 4
( sqrt( beta/ alpha ) + sqrt( alpha / beta ) ) ^2
= beta / alpha + alpha / beta + 2
= (alpha^2 + beta^2 + 2 alpha beta ) / (alpha . beta)
= (alpha + beta)^2 / ( alpha . beta)
= (25 / 4 ) / ( 1 /2)
= 25 / 2
2) a, b 是方程 x^2 + 5x + 2 = 0 的两根
韦达定理: a+ b = -5, ab = 2
==》 1/a^2 + 1/b^2 = (a^2 + b^2)/ (ab)^2 =( (a+b)^2 - 2ab ) / (ab)^2 = 21/ 4
1/a^2 . 1/ b^2 = 1/ 4
所以,1/a^2 , 1/b^2 是方程 x^2 - 21/4 x + 1/4 = 0 的两根
3) 设两根为 a,b 根据韦达定理,
a+ b = -k - 1
ab =k+3
==》 a^2 + b^2 = (a+b)^2 - 2ab = (k+1)^2 - 2(k+3) = k^2 -5 = 4
==> k = 3 ou -3
因为 原方程判别式为 delta = (k+1)^2 - 4 (k+3) = (k-1)^2 - 12
若 k = 3 则 delta = ( 3-1)^2 - 12 = 4 - 12 <0, 故舍去此值
若k = - 3,则 delta =(-3-1)^2 -12 = 16-12 >0
综上所属, k = -3
alpha + beta = - 5/2
alpha . beta = 1 / 2
==》
alpha^2 + alpha beta + beta^2
= (alpha + beta)^2 - alpha beta = 25/4 - 1/2 = 23 / 4
( sqrt( beta/ alpha ) + sqrt( alpha / beta ) ) ^2
= beta / alpha + alpha / beta + 2
= (alpha^2 + beta^2 + 2 alpha beta ) / (alpha . beta)
= (alpha + beta)^2 / ( alpha . beta)
= (25 / 4 ) / ( 1 /2)
= 25 / 2
2) a, b 是方程 x^2 + 5x + 2 = 0 的两根
韦达定理: a+ b = -5, ab = 2
==》 1/a^2 + 1/b^2 = (a^2 + b^2)/ (ab)^2 =( (a+b)^2 - 2ab ) / (ab)^2 = 21/ 4
1/a^2 . 1/ b^2 = 1/ 4
所以,1/a^2 , 1/b^2 是方程 x^2 - 21/4 x + 1/4 = 0 的两根
3) 设两根为 a,b 根据韦达定理,
a+ b = -k - 1
ab =k+3
==》 a^2 + b^2 = (a+b)^2 - 2ab = (k+1)^2 - 2(k+3) = k^2 -5 = 4
==> k = 3 ou -3
因为 原方程判别式为 delta = (k+1)^2 - 4 (k+3) = (k-1)^2 - 12
若 k = 3 则 delta = ( 3-1)^2 - 12 = 4 - 12 <0, 故舍去此值
若k = - 3,则 delta =(-3-1)^2 -12 = 16-12 >0
综上所属, k = -3
追问
这是什么鬼啊
看不懂
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询