1)已知y=(√1-2x+x^2)+(√x^2-4x+4)+(√4x^2+4x+1),试求使y的值恒等于常数的x的取值范围2)已知(4√x-1)+(6√y-2)-10=x+y,求(2x-y)^2011的值3)已知x=0.44,求二次根式√1-x-x^2+x^3的值限时3小
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![1)已知y=(√1-2x+x^2)+(√x^2-4x+4)+(√4x^2+4x+1),试求使y的值恒等于常数的x的取值范围2)已知(4√x-1)+(6√y-2)-10=x+y,求(2x-y)^2011的值3)已知x=0.44,求二次根式√1-x-x^2+x^3的值限时3小](/uploads/image/z/11480798-38-8.jpg?t=1%EF%BC%89%E5%B7%B2%E7%9F%A5y%3D%EF%BC%88%E2%88%9A1-2x%2Bx%5E2%EF%BC%89%2B%EF%BC%88%E2%88%9Ax%5E2-4x%2B4%EF%BC%89%2B%EF%BC%88%E2%88%9A4x%5E2%2B4x%2B1%EF%BC%89%2C%E8%AF%95%E6%B1%82%E4%BD%BFy%E7%9A%84%E5%80%BC%E6%81%92%E7%AD%89%E4%BA%8E%E5%B8%B8%E6%95%B0%E7%9A%84x%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B42%EF%BC%89%E5%B7%B2%E7%9F%A5%EF%BC%884%E2%88%9Ax-1%EF%BC%89%2B%EF%BC%886%E2%88%9Ay-2%EF%BC%89-10%3Dx%2By%2C%E6%B1%82%EF%BC%882x-y%EF%BC%89%5E2011%E7%9A%84%E5%80%BC3%EF%BC%89%E5%B7%B2%E7%9F%A5x%3D0.44%2C%E6%B1%82%E4%BA%8C%E6%AC%A1%E6%A0%B9%E5%BC%8F%E2%88%9A1-x-x%5E2%2Bx%5E3%E7%9A%84%E5%80%BC%E9%99%90%E6%97%B63%E5%B0%8F)
1)已知y=(√1-2x+x^2)+(√x^2-4x+4)+(√4x^2+4x+1),试求使y的值恒等于常数的x的取值范围2)已知(4√x-1)+(6√y-2)-10=x+y,求(2x-y)^2011的值3)已知x=0.44,求二次根式√1-x-x^2+x^3的值限时3小
1)已知y=(√1-2x+x^2)+(√x^2-4x+4)+(√4x^2+4x+1),试求使y的值恒等于常数的x的取值范围
2)已知(4√x-1)+(6√y-2)-10=x+y,求(2x-y)^2011的值
3)已知x=0.44,求二次根式√1-x-x^2+x^3的值
限时3小时,答得越快越好
1)已知y=(√1-2x+x^2)+(√x^2-4x+4)+(√4x^2+4x+1),试求使y的值恒等于常数的x的取值范围2)已知(4√x-1)+(6√y-2)-10=x+y,求(2x-y)^2011的值3)已知x=0.44,求二次根式√1-x-x^2+x^3的值限时3小
①Y=√(X-1)²+√(X-2)²+√(2X-1)²
由题意,式中不含未知项
∴当且仅当X-1≤0和X-2≤0且√2X-1≥0时,Y恒等于常数
解得0.5≤X≤1
②凑配法:移项,全部移到左面得(4√x-1)+(6√y-2)-10-X-Y=0即X+Y-(4√x-1)-(6√y-2)+10=0
X-1-(4√x-1)+Y-2-(6√y-2)+13=0
√(X-1)²-(4√X-1)+4+√(Y-2)²-(6√Y-2)+9=0
{√(X-1)-2}²+{√(Y-2)-3}²=0
所以√(X-1)-2=0且√(Y-2)-3=0
解得X=5,Y=11
所以所求=(2×5-11)^2011=(-1)^2011=-1
③原式=√X²(X-1)-(X-1)=√(X²-1)(X-1)=√(X-1)²(X-1)=(1-X)√X+1=0.56×√1.44=0.56×1.2=0.672
我自信的人为我的理解是正确的,合作双赢!