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解方程; (-3x)^2-(2x+1)(3x-2)-3(x+2)(x-2)=0 计算 1.已知x^2+2x+y^2-6y+10=0,求x,y的值
解方程;
(-3x)^2-(2x+1)(3x-2)-3(x+2)(x-2)=0
计算
1.已知x^2+2x+y^2-6y+10=0,求x,y的值
2.(1-2^2分之1)(1-3^2分之1)(1-4^2分之1).(1-99^2分之1)(1-100^2分之1)
人气:305 ℃ 时间:2020-05-02 14:48:51
解答
(-3x)^2-(2x+1)(3x-2)-3(x+2)(x-2)=0
9x²-(6x²-x-2)-3(x²-4)=0
9x²-6x²+x+2-3x²+12=0
x+14=0
x=-14
1.已知x^2+2x+y^2-6y+10=0
x^2+2x+1+y^2-6y+9=0
(x+1)²+(y-3)²=0
x+1=0
y-3=0
x=-1 ,y=3
2.(1-2^2分之1)(1-3^2分之1)(1-4^2分之1).(1-99^2分之1)(1-100^2分之1)
=(1+ 1/2)(1- 1/2)(1+1/3)(1-1/3)*.*(1+1/100)(1-1/100)
=(1+ 1/2)(1+1/3)*...(1+1/100)*(1- 1/2)(1-1/3)*.*(1-1/100)
=101/2 * 1/100
=101/200
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