If x 2, when x, x 1 x 2 has a minimum value of

x+1 (x-2)=x-2+1 (x-2) +2 When x-2=1 (x-2), i.e., x=3, there is a minimum value that substitutes x=3 into 3+1 (3-2)=4

2 (x +2) (x +1) (x +2) (x +1)=[(x +1)+1] (x +1)>=since a +b >=2ab

2 (x +1) (x +1) =2 so the minimum value is 2

x+1 (x-2)=(x-2)+1 (x-2)+2Because x>2, x-2>0

Original formula = (2 root number (x-2) * 1 (x-20)) + 2 = 4 (if and only if (x-2) = 1 (x-2) (i.e. x=3) "=" holds).

The key to this problem is to match the form of the mean inequality to the conditions under which the mean inequality applies.

x>1 so x-1>0

So x+1 (x-1)=(x-1)+1 (x-1)+1>=1+2*root number [(x-1)*1 (x-1)]=1+2

When x-1=1 (x-1), an equal sign is taken.

x-1)^2=1

x-1>0

x=2 so the equal sign can be taken.

So x+1 (x-1)>=1+2

So the minimum value = 3

f(x)=(x²-2x+2)/(x-1)

x-1)^2+1/(x-1)

x-1+1/(x-1)

2. When the file mu and only when x=2, the fissure=

Hope, unbridled imitation thank you.

y=(x^2-2x+2)/(2x-2)

(x-1)^2+1]/2(x-1)

The numerator and denominator are divided by x-1).

(x-1)+1/(x-1)]/2

x＞1,x-1>0

y>=2*(1 2) (x-1)*[1 (x-1)] (equal sign if and only if x-1=1 (x-1)) ymin = 1

Answer: y=(x 2-x+1) (x-1)=[x(x-1)+1] (x-1).

x+1/(x-1)=x-1

1/(x-1)

Since x>1, then x-1>0

So y=x-1

1/(x-1)

That is, the minimum value is 2Complete.

Solution: This function can be reduced to:

y=x+1+4/(x+1)+5

Let t=x+1, obviously t>0

That is: y=t+4 t+5

Make use of the mean theorem.

y>=2*2+5=9 (equal sign if and only if x=2) i.e. ymin=9

1,2+3x+4 x=2+(3x+4 x) 2+4 3 [when 3x=4 x,x>0, take the = sign], that is, the minimum value is 2+4 3

2, SK=A1(1-Q) K 1-Q=A1-AK Q 1-Q, generation number, AK=243

The root of 3,5x -7x-6=0 is x1=2 [rounded], x2=3 5, so the cosine value of the included angle is 3 5, the included angle is an acute angle, and its sine value is 1-(3 5) =, then the area is (3 5

(1) According to 3x+4 x is greater than or equal to 3x*4 x under the root number, you can get the answer: 2 plus 4 root number 3

2) According to the formula: n=6, so an=243

3) From the known: cosa=-3 5, so sina = 4 5, the area is 6

Solution: This function can be reduced to:

y=x+1+4/(x+1)+5

Let t=x+1, obviously t>0

That is: y=t+4 t+5

Make use of the mean theorem.

y>=2*2+5=9 (equal sign if and only if x=2) i.e. ymin=9

Original = (x 2-1) (x-1)+1 (x-1)(x+1)+1 (x-1).

x-1)+1/(x-1)+2

Because x>1, x-1>0

So the original formula = 2 + 2 = 4

So the minimum value of the original formula is 4, and the minimum value is taken if and only if x=2.