解方程√(x²-1)+√(x²+4x+3)=√(3x²+4x+1)
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解方程√(x²-1)+√(x²+4x+3)=√(3x²+4x+1)
解方程√(x²-1)+√(x²+4x+3)=√(3x²+4x+1)
解方程√(x²-1)+√(x²+4x+3)=√(3x²+4x+1)
√(x²-1)+√(x²+4x+3)=√(3x²+4x+1)
两边平方得:
x²-1+2*√[(x²-1)(x²+4x+3)]+x²+4x+3=3x²+4x+1
2*√[(x²-1)(x²+4x+3)]=x²-1
(x²-1)(x²+4x+3)=(x²-1)²
(x²-1)(x²+4x+3)-(x²-1)²=0
(x²-1)(x²+4x+3-x²+1)=0
(x-1)(x+1)(4x+4)=0
∴x-1=0 x+1=0
∴x1=1
x2=-1
可以用集合求解.
1,(x²-1)>=0;
2,(x²+4x+3)>=0;
3,(3x²+4x+1)>=0
解出来的X的值的范围用集合来表示.
-1
两边平方,再化简。
最后得:
X1=1
X1=-1
X3=-13/3
x²-1+2*√[(x²-1)(x²+4x+3)]+x²+4x+3=3x²+4x+1
2*√[(x²-1)(x²+4x+3)]=x²-1
4*(x²-1)(x²+4x+3)=(x²-1)² 前面那位这步左边少乘了个4
(x²...
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x²-1+2*√[(x²-1)(x²+4x+3)]+x²+4x+3=3x²+4x+1
2*√[(x²-1)(x²+4x+3)]=x²-1
4*(x²-1)(x²+4x+3)=(x²-1)² 前面那位这步左边少乘了个4
(x²-1)(4x²+16x+12)-(x²-1)²=0
(x-1)(x+1)(3x²+16x+13)=0
(x+1)²(x-1)(3x+13)=0
x1=x2=-1,x3=1,x4=-13/3
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