If the equation y y 2 y 1 y 2 0, then y

Remove parentheses: y -2y + y + y - 2 = 0

Simplification: 2y -2y-2 = 0

y²-y-1=0

Method 1: Recipe: y -y + (1 2) = 5 4y - 1 2) = 5 4

y-1 2= root number 5 2

y= (root number 5+1) 2

Method 2: Formula method: y=(-b root number b -4ac) 2ay=(1 root number 5) 2

The answer is the same for both sides, whichever method you choose.

From the question, it can be seen that y(y+2)=0; y-1)(y+2)=0 solution y=0;y=-2

The solution is y=1;y=-2

To sum up: y=-2

y(y-2)+(y-1)(y+2)=0

y squared - 2y + y square + 2y - y - 2 = 0

2y square = 2

y squared 1y plus or minus 1

x=1 is substituted into the equation to obtain the jujube: y-1=0, and the number of stools on both sides of the equation is derived from x: 2xy+x 2y'+2yy'lnx+y^2/x=0

Substituting x=1, y=1 to the above formula: 2+y'Biyu +1=0, get: y'=-3 is y'|(x=1)=-3

2 (y bright sky high loss 1) 3 (y a 1), 2y 2 3y a respectful ruler 3, solution: y 5.

It can be obtained according to the meaning of the title.

A 2 5, so there is: A 3.

For equation 2 (xy)=x+y, the number of bridges on both sides is derived.

2^(xy) ln2 (y+xy' )1+y'

The solution is complete. y' =1-y2^(xy) ln2 ] x 2^(xy) ln2 -1 ].

b.Just substitute the formula directly.

The characteristic equation r +2r +1 = 0

r+1)²=0

r1,2=-1

This is the second case in the diagram.

y=(c1+c2x)e^(-x)

Eligible is B

b.Directly substitute the formula, that is, the trouser base calendar.

The characteristic equation r is +2r+1=0

r+1)²=0

r1,2=-1

This is the second case in the diagram.

y=(c1+c2x)e^(-x)

Those who meet the conditions of Fengsong are b

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