Với Giá Trị Nào Của M Thì Hai đường Thẳng D1:(m−3)x+2y+m2−1=0 ...

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Câu hỏi

Với giá trị nào của m thì hai đường thẳng d1:m−3x+2y+m2−1=0 và d2:−x+my+m2−2m+1=0 cắt nhau? A. m≠1 . B. m≠1m≠2 . C. m≠2 . D. m≠1m≠2 . Đáp án và lời giải Đáp án:B Lời giải:Lời giải d1:m−3x+2y+m2−1=0d2:−x+my+m2−2m+1=0 →d1∩d2=Mm=0→d1:−3x+2y−1=0d2:−x+1=0→thoaû maõnm=0→m−3−1=2m⇔m=1m=2. Chọn B

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