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> 已知关于x的一元二次方程mx2-(2m-1)x+m-2=0(m>0)(1)求证:这个方程有两个不相等的实数根;(2)如果这个方程的两个实数根分别为x1,x2,且(x1-3)(x2-3)=5m,求m的值
已知关于x的一元二次方程mx2-(2m-1)x+m-2=0(m>0)(1)求证:这个方程有两个不相等的实数根;(2)如果这个方程的两个实数根分别为x1,x2,且(x1-3)(x2-3)=5m,求m的值
题目简介
已知关于x的一元二次方程mx2-(2m-1)x+m-2=0(m>0)(1)求证:这个方程有两个不相等的实数根;(2)如果这个方程的两个实数根分别为x1,x2,且(x1-3)(x2-3)=5m,求m的值
题目详情
已知关于x的一元二次方程mx
2
-(2m-1)x+m-2=0(m>0)
(1)求证:这个方程有两个不相等的实数根;
(2)如果这个方程的两个实数根分别为x
1
,x
2
,且(x
1
-3)(x
2
-3)=5m,求m的值.
题型:解答题
难度:中档
来源:上海
答案
(1)判别式△=(2m-1)2-4m(m-2)
=4m2-4m+1-4m2+8m
=4m+1
∵m>0
∴4m+1>0
所以方程有两个不相等的实数根.
(2)由韦达定理得
x1+x2=
class="stub"2m-1
m
x1x2=
class="stub"m-2
m
所以(x1-3)(x2-3)=5m
x1x2-3(x1+x2)+9=5m
class="stub"m-2
m
-3×
class="stub"2m-1
m
+9=5m
两边同时乘以m并化简
m-2-6m+3+9m=5m2
5m2-4m-1=0
(5m+1)(m-1)=0
解得m=1或m=-
class="stub"1
5
(舍去)
经检验m=1是方程的根.
所以m的值是1.
上一篇 :
(10分)先化简,再求值:,其中.-八年
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The old woman was counting t
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题目简介
已知关于x的一元二次方程mx2-(2m-1)x+m-2=0(m>0)(1)求证:这个方程有两个不相等的实数根;(2)如果这个方程的两个实数根分别为x1,x2,且(x1-3)(x2-3)=5m,求m的值
题目详情
(1)求证:这个方程有两个不相等的实数根;
(2)如果这个方程的两个实数根分别为x1,x2,且(x1-3)(x2-3)=5m,求m的值.
答案
=4m2-4m+1-4m2+8m
=4m+1
∵m>0
∴4m+1>0
所以方程有两个不相等的实数根.
(2)由韦达定理得
x1+x2=
x1x2=
所以(x1-3)(x2-3)=5m
x1x2-3(x1+x2)+9=5m
两边同时乘以m并化简
m-2-6m+3+9m=5m2
5m2-4m-1=0
(5m+1)(m-1)=0
解得m=1或m=-
经检验m=1是方程的根.
所以m的值是1.