1.已知a+b=5,ab=3,求a²+b²的值.(提示:利用公式(a+b)²=a²+2ab+b²).2.解不等式(2x-5)²+(3x+1)²>13(x²-10).
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![1.已知a+b=5,ab=3,求a²+b²的值.(提示:利用公式(a+b)²=a²+2ab+b²).2.解不等式(2x-5)²+(3x+1)²>13(x²-10).](/uploads/image/z/8668113-33-3.jpg?t=1.%E5%B7%B2%E7%9F%A5a%2Bb%3D5%2Cab%3D3%2C%E6%B1%82a%26%23178%3B%2Bb%26%23178%3B%E7%9A%84%E5%80%BC.%EF%BC%88%E6%8F%90%E7%A4%BA%EF%BC%9A%E5%88%A9%E7%94%A8%E5%85%AC%E5%BC%8F%EF%BC%88a%2Bb%29%26%23178%3B%3Da%26%23178%3B%2B2ab%2Bb%26%23178%3B%29.2.%E8%A7%A3%E4%B8%8D%E7%AD%89%E5%BC%8F%282x-5%29%26%23178%3B%2B%283x%2B1%29%26%23178%3B%3E13%28x%26%23178%3B-10%29.)
1.已知a+b=5,ab=3,求a²+b²的值.(提示:利用公式(a+b)²=a²+2ab+b²).2.解不等式(2x-5)²+(3x+1)²>13(x²-10).
1.已知a+b=5,ab=3,求a²+b²的值.(提示:利用公式(a+b)²=a²+2ab+b²).
2.解不等式(2x-5)²+(3x+1)²>13(x²-10).
1.已知a+b=5,ab=3,求a²+b²的值.(提示:利用公式(a+b)²=a²+2ab+b²).2.解不等式(2x-5)²+(3x+1)²>13(x²-10).
1.∵a+b=3,
∴a2+2ab+b2=9,
∵a2+b2=5,
∴2ab=9-5=4.
解得ab=2.
故答案为:2.
2.解4x²-20x+25+9x²+6x+1>13x²-130
-14x>-156
x
a+b)²=a²+2ab+b²,,,25-3x2=19