已知a≠b,且a,b满足2a 紧急数学题.已知a≠b,且a,b满足2a²+4a-3=0,2b²+4b-3=0.求下列各式的值.(1)(a-1)(b-1) (2)b/a+a/b
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![已知a≠b,且a,b满足2a 紧急数学题.已知a≠b,且a,b满足2a²+4a-3=0,2b²+4b-3=0.求下列各式的值.(1)(a-1)(b-1) (2)b/a+a/b](/uploads/image/z/1057842-18-2.jpg?t=%E5%B7%B2%E7%9F%A5a%E2%89%A0b%2C%E4%B8%94a%2Cb%E6%BB%A1%E8%B6%B32a+%E7%B4%A7%E6%80%A5%E6%95%B0%E5%AD%A6%E9%A2%98.%E5%B7%B2%E7%9F%A5a%E2%89%A0b%2C%E4%B8%94a%2Cb%E6%BB%A1%E8%B6%B32a%26sup2%3B%2B4a-3%3D0%2C2b%26sup2%3B%2B4b-3%3D0.%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E7%9A%84%E5%80%BC.%EF%BC%881%EF%BC%89%28a-1%29%28b-1%29+%EF%BC%882%EF%BC%89b%2Fa%2Ba%2Fb)
已知a≠b,且a,b满足2a 紧急数学题.已知a≠b,且a,b满足2a²+4a-3=0,2b²+4b-3=0.求下列各式的值.(1)(a-1)(b-1) (2)b/a+a/b
已知a≠b,且a,b满足2a 紧急数学题.
已知a≠b,且a,b满足2a²+4a-3=0,2b²+4b-3=0.求下列各式的值.
(1)(a-1)(b-1)
(2)b/a+a/b
已知a≠b,且a,b满足2a 紧急数学题.已知a≠b,且a,b满足2a²+4a-3=0,2b²+4b-3=0.求下列各式的值.(1)(a-1)(b-1) (2)b/a+a/b
2a²+4a-3=0,
2b²+4b-3=0.
说明a,b是方程2x^2+4x-3=0的两根
因此有
a+b=-2
ab=-3/2
(1)
(a-1)(b-1)
=ab-(a+b)+1
=-3/2+2+1
=3/2
(2)
b/a+a/b
=(a^2+b^2)/ab
=((a+b)^2-2ab)/ab
=(4-2*(-3/2))/(-3/2)
=-14/3
由韦达定理a+b=-2,ab=-3/2,
(a-1)(b-1)=ab-(a+b)+1
b/a+a/b =[(a+b)^2-2ab]/ab
a≠b,且a,b满足2a²+4a-3=0, 2b²+4b-3=0
即a,b是关于2x^2+4x-3=0的解
由韦达定理得
a+b=-2
ab=-3/2
(1)(a-1)(b-1)
=ab-(a+b)+1
=-3/2-(-2)+1
=3-3/2
=3/2
(2)b/a+a/...
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a≠b,且a,b满足2a²+4a-3=0, 2b²+4b-3=0
即a,b是关于2x^2+4x-3=0的解
由韦达定理得
a+b=-2
ab=-3/2
(1)(a-1)(b-1)
=ab-(a+b)+1
=-3/2-(-2)+1
=3-3/2
=3/2
(2)b/a+a/b
=(a^2+b^2)/ab
=[(a+b)^2-2ab]/ab
=(a+b)^2/ab-2
=4/(-3/2)-2
=4*(-2/3)-2
=-8/3-2
=-14/3
收起
可以把ab看成2x²+4x-3=0 的2个根
a+b=-2
ab=-3/2
(1)(a-1)(b-1)=ab-(a+b)+1=-3/2+2+1=3/2
(2)b/a+a/b =a^2+b^2/ab=[(a+b)^2-2ab]/ab=-14/3
a≠b,且a,b满足2a²+4a-3=0, 2b²+4b-3=0.
所以a、b是方程x^2+4x-3=0的两根 a+b=-2 ab=-3/2
(1))(a-1)(b-1) =ab-(a+b)+1=-3/2+2+1=3/2
(2)b/a+a/b =(a+b)/ab=4/3