若方程x²-(3m+2)x+3m-1=0的两根互为相反数,那么这两根是 因式分解:x²-2x-2y²+1
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若方程x²-(3m+2)x+3m-1=0的两根互为相反数,那么这两根是 因式分解:x²-2x-2y²+1
若方程x²-(3m+2)x+3m-1=0的两根互为相反数,那么这两根是 因式分解:x²-2x-2y²+1
若方程x²-(3m+2)x+3m-1=0的两根互为相反数,那么这两根是 因式分解:x²-2x-2y²+1
x^2-(3m+2)x+3m-1=0的两根之和为0
即3m+2=0
3m=-2
原式为:x^2-3=0
两根是 x=√3或x=-√3
x²-2x-2y²+1
=x²-2x+1-2y²
=(x-1)²-2y²
=(x-1+√2y)(x-1-√2y)
题意不明
两根互为相反数,所以3m+2=0,m=-2/3
所以方程为x^2-3=0
两根为 根号3,-根号3
x^2-2x+1-2y^2
=(x-1)^2-2y^2
=(x+根号2y-1)(x-根号2y-1)