2x²-7x+6=0; 已知x²-5x=3,求(x-1)(2x-1)-(x+1)²+1的值
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![2x²-7x+6=0; 已知x²-5x=3,求(x-1)(2x-1)-(x+1)²+1的值](/uploads/image/z/3155109-69-9.jpg?t=2x%26%23178%3B-7x%2B6%3D0%EF%BC%9B+%E5%B7%B2%E7%9F%A5x%26%23178%3B-5x%3D3%2C%E6%B1%82%EF%BC%88x-1%EF%BC%89%EF%BC%882x-1%EF%BC%89-%EF%BC%88x%2B1%EF%BC%89%26%23178%3B%2B1%E7%9A%84%E5%80%BC)
2x²-7x+6=0; 已知x²-5x=3,求(x-1)(2x-1)-(x+1)²+1的值
2x²-7x+6=0; 已知x²-5x=3,求(x-1)(2x-1)-(x+1)²+1的值
2x²-7x+6=0; 已知x²-5x=3,求(x-1)(2x-1)-(x+1)²+1的值
1.2x²-7x+6=0 (2x-3)(x-2)=0 x=2或2/3
2.已知x²-5x=3
(x-1)(2x-1)-(x+1)²+1
=(2x²-3x+1)-(x²+2x+1)+1
=2x²-3x+1-x²-2x-1+1
=x²-5x+1
=3+1
=4
2x²-10x-6=0
2x²-7x+6=0
相减
-3x-12=0
x=-4
(x-1)(2x-1)-(x+1)
=-5*(-9)-(-4+1)²+1
=37