因式分解题(初二)(1)(x²+y²)²-4x²y²还有一题很难,已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,求y的值
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![因式分解题(初二)(1)(x²+y²)²-4x²y²还有一题很难,已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,求y的值](/uploads/image/z/1167551-71-1.jpg?t=%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E9%A2%98%EF%BC%88%E5%88%9D%E4%BA%8C%EF%BC%89%EF%BC%881%EF%BC%89%EF%BC%88x%26%23178%3B%2By%26%23178%3B%EF%BC%89%26%23178%3B-4x%26%23178%3By%26%23178%3B%E8%BF%98%E6%9C%89%E4%B8%80%E9%A2%98%E5%BE%88%E9%9A%BE%2C%E5%B7%B2%E7%9F%A5P%3D3xy-8x%2B1%2CQ%3Dx-2xy-2%2C%E5%BD%93x%E2%89%A00%E6%97%B6%2C3P-2Q%3D7%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82y%E7%9A%84%E5%80%BC)
因式分解题(初二)(1)(x²+y²)²-4x²y²还有一题很难,已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,求y的值
因式分解题(初二)
(1)(x²+y²)²-4x²y²
还有一题很难,
已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,求y的值
因式分解题(初二)(1)(x²+y²)²-4x²y²还有一题很难,已知P=3xy-8x+1,Q=x-2xy-2,当x≠0时,3P-2Q=7恒成立,求y的值
解
(x²+y²)²-4x²y²
=(x²+y²)²-(2xy)²
=(x²+y²-2xy)(x²+y²+2xy)
=(x-y)²(x+y)²
3P-2Q=7
即3(3xy-8x+1)-2(x-2xy-2)=7
即(9xy+4xy)+(-24x-2x)+(3+4)=7
即13xy-26x=0
∵x≠0
∴两边除以x得:13y-26=0
∴y=2