已知-½≤x≤1,则√(x-1)²+(x-1)的绝对值+√(4x²+4x+1)的值为?
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![已知-½≤x≤1,则√(x-1)²+(x-1)的绝对值+√(4x²+4x+1)的值为?](/uploads/image/z/2589353-17-3.jpg?t=%E5%B7%B2%E7%9F%A5-%26%23189%3B%E2%89%A4x%E2%89%A41%2C%E5%88%99%E2%88%9A%EF%BC%88x-1%29%26%23178%3B%2B%28x-1%29%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%2B%E2%88%9A%284x%26%23178%3B%2B4x%2B1%29%E7%9A%84%E5%80%BC%E4%B8%BA%3F)
已知-½≤x≤1,则√(x-1)²+(x-1)的绝对值+√(4x²+4x+1)的值为?
已知-½≤x≤1,则√(x-1)²+(x-1)的绝对值+√(4x²+4x+1)的值为?
已知-½≤x≤1,则√(x-1)²+(x-1)的绝对值+√(4x²+4x+1)的值为?
因为:-½≤x≤1
所以 √(x - 1)² + | x - 1| + √(4x² + 4x + 1)
= 1 - x + 1 - x + √(2x + 1)²
= 2 - 2x + (2x + 1)
= 2 - 2x + 2x + 1
= 3