22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=
22222请问这个题目怎么做解:2+2-√(2x2)=2+2-√(4)=2+2-2=22+√(2x2)-2=2+√(4)-2=2+2-2=22÷2+2÷2=1+1=2=42,x222x32求过程(x-2)2=(2x+3)2(x-2)2-(2x+3)2=0[(x-2)+(2x+3)][(x-2)-(2x+3)]=0(3x+1)(-x-5)=0x=-1/3orx=-5(X-2)2=(2x+3)2.(X-2)2-(2x+3)2=0(x-2+2x+3)(x-2-2x-3)=0(3x+1)(-x-5)=0x1=