1.方程x²=0的实数根的个数为( )A.0个 B.1个 C.2个 D.无数个2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( )解方程:(1)(x+2)²-8=0 (2)x(x-3)=x (3)(x-1)²/x²-(x-1)/x
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![1.方程x²=0的实数根的个数为( )A.0个 B.1个 C.2个 D.无数个2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( )解方程:(1)(x+2)²-8=0 (2)x(x-3)=x (3)(x-1)²/x²-(x-1)/x](/uploads/image/z/170713-1-3.jpg?t=1.%E6%96%B9%E7%A8%8Bx%26%23178%3B%3D0%E7%9A%84%E5%AE%9E%E6%95%B0%E6%A0%B9%E7%9A%84%E4%B8%AA%E6%95%B0%E4%B8%BA%EF%BC%88+%EF%BC%89A.0%E4%B8%AA+B.1%E4%B8%AA+C.2%E4%B8%AA+D.%E6%97%A0%E6%95%B0%E4%B8%AA2.%E5%A6%82%E5%9B%BE%2C%E2%96%B3ABC%E7%9A%84%E9%A1%B6%E7%82%B9%E9%83%BD%E5%9C%A8%E5%9C%86O%E4%B8%8A%2C%E5%B7%B2%E7%9F%A5%E2%88%A0B%3D60%C2%B0%2C%E5%88%99%E2%88%A0CAO%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%88++%EF%BC%89%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%9A%EF%BC%881%EF%BC%89%28x%2B2%29%26%23178%3B-8%3D0++%EF%BC%882%EF%BC%89x%28x-3%29%3Dx+%EF%BC%883%EF%BC%89%28x-1%29%26%23178%3B%2Fx%26%23178%3B-%EF%BC%88x-1%EF%BC%89%2Fx)
1.方程x²=0的实数根的个数为( )A.0个 B.1个 C.2个 D.无数个2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( )解方程:(1)(x+2)²-8=0 (2)x(x-3)=x (3)(x-1)²/x²-(x-1)/x
1.方程x²=0的实数根的个数为( )A.0个 B.1个 C.2个 D.无数个
2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( )
解方程:(1)(x+2)²-8=0 (2)x(x-3)=x (3)(x-1)²/x²-(x-1)/x-(2)=0
1.方程x²=0的实数根的个数为( )A.0个 B.1个 C.2个 D.无数个2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( )解方程:(1)(x+2)²-8=0 (2)x(x-3)=x (3)(x-1)²/x²-(x-1)/x
方程x²=0的实数根的个数为( C)A.0个 B.1个 C.2个 D.无数个
为两个相等的实数根
2.如图,△ABC的顶点都在圆O上,已知∠B=60°,则∠CAO的度数( 30º)
∠AOC=2∠B=120º
∵∠CAO=∠ACO
∴∠CAO=(180º-120º)÷2=30º
解方程:
(1)(x+2)²-8=0
(x+2)²=8
x+2=±2√2
x1=-2+2√2 x2=-2-2√2
(2)x(x-3)=x
x(x-3)-x =0
x(x-3-1)=0
x1=0 x2=4
(3)(x-1)²/x²-(x-1)/x-(2)=0
设(x-1)/x=t,则
t²-t-2=0
(t-2)(t+1)=0
t1=2 t2=-1
于是
(x-1)/x=2时:
x-1=2x
x=-1
(x-1)/x=-1时:
x-1=-x
2x=1
x=1/2
所以x1=-1 x2=1/2