已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为(x1,y1)和(x2,y2),且y1+y2=4(x1+x2),求k的值.
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![已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为(x1,y1)和(x2,y2),且y1+y2=4(x1+x2),求k的值.](/uploads/image/z/8570594-2-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8B%E7%BB%84x%2Bky%3D3%3B2x%26%23178%3B%2By%26%23178%3B%3D6%28k%3E1%29%E7%9A%84%E4%B8%A4%E7%BB%84%E5%AE%9E%E6%95%B0%E8%A7%A3%E4%B8%BA%EF%BC%88x1%2Cy1%29%E5%92%8C%EF%BC%88x2%2Cy2%29%2C%E4%B8%94y1%2By2%3D4%28x1%2Bx2%29%2C%E6%B1%82k%E7%9A%84%E5%80%BC.)
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已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为(x1,y1)和(x2,y2),且y1+y2=4(x1+x2),求k的值.
已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为(x1,y1)和(x2,y2),且y1+y2=4(x1+x2),求k的值.
已知方程组x+ky=3;2x²+y²=6(k>1)的两组实数解为(x1,y1)和(x2,y2),且y1+y2=4(x1+x2),求k的值.
①得,x=3-ky③
代入②得,18-12ky+2k²y²+y²=6
(2k²+1)y²-12ky+12=0
∵两组实数解为(x1,y1)和(x2,y2),
∴y1+y2=12/(2k²+1) x1+x2=6-k(y1+y2)=6-12k/(2k²+1)=(12k²-12k+6)/(2k²+1)
∵y1+y2=4(x1+x2),
∴12/(2k²+1)=4(12k²-12k+6)/(2k²+1)
∴2k²-2k=0
∴k1=0 k2=1
∵⊿=(-12k)²-4(2k²+1)≥0
∴k=1