英语翻译若函数y=根号下[（a²-1）x²+(a-1)x+2/（a+1）]的定义域为R,求实数a的取值范围.①a²-1=0,即当a=1时,不等式为2/2=1>=0恒成立,所以a=1符合题意.②当a²-1≠0时必须……（不好打,省

英语翻译若函数y=根号下[（a²-1）x²+(a-1)x+2/（a+1）]的定义域为R,求实数a的取值范围.①a²-1=0,即当a=1时,不等式为2/2=1>=0恒成立,所以a=1符合题意.②当a²-1≠0时必须……（不好打,省
英语翻译
若函数y=根号下[（a²-1）x²+(a-1)x+2/（a+1）]的定义域为R,求实数a的取值范围.
①a²-1=0,即当a=1时,不等式为2/2=1>=0恒成立,所以a=1符合题意.
②当a²-1≠0时必须……（不好打,省略了^-^）
综上所述,所求a的范围是[1,9]

英语翻译若函数y=根号下[（a²-1）x²+(a-1)x+2/（a+1）]的定义域为R,求实数a的取值范围.①a²-1=0,即当a=1时,不等式为2/2=1>=0恒成立,所以a=1符合题意.②当a²-1≠0时必须……（不好打,省
If the function y = root under [(a ² -1) x ² + (a-1) x +2 / (a ​​+1)] of the domain of R,the number of a realistic range of values.
Solution:① a ² -1 = 0,that is,when a = 1,the inequality 2 / 2 = 1> = 0 always true,so a = 1 line meaning of the questions.
② When a ² -1 ≠ 0 must ...seek a range of [1,9]

If the function y = radical signs [ ( a L - 1 ) x L + ( A-1 ) x+2 / ( a+1 ) ] domain is R, seeking a subset.Solution : a L - 1 = 0, i.e., when a = 1, 2 / 2 = 1 inequalities of > = 0 founded, so filled with a = 1.While the a L - 1 and 0 must be ... ... ( not play, omitting the ^ - ^ )To sum up, ask a is in the range [1, 9]

If the function under the square y = [(a 2-1) x 2 + (a) x + 2 / (a + 1)] the domain for R, realistic and number of a value range.
Solution: (1) a 2-1 = 0, that is, when a = 1, inequality for 2/2 =...

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If the function under the square y = [(a 2-1) x 2 + (a) x + 2 / (a + 1)] the domain for R, realistic and number of a value range.
Solution: (1) a 2-1 = 0, that is, when a = 1, inequality for 2/2 = 1 > = 0 constant set up, so a = 1 meet cet4.
(2) when a 2-1 indicates a 0 must... (not good play, bypassing a ^-^)
To sum up, but for a range is [1, 9]

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