1 Knowing a b 3, ab 1, find the value of 1 a 1 b.

Solution:1. Because a+b=3, ab=1, 1 a+1 b=(a+b) ab

2。Because 1 a+1 b=5

So (a+b) ab=5

a+b=5ab

So (2a+ab+2b) (a--2ab+b)[2(a+b)+ab] [(a+b)--2ab](10ab+ab) (5ab--2ab).

11ab/3ab

3。Because x 2--4x+1=0, then x--4+1 x=0x+1 x=4

x^2+2+1/x^2=16

x^2+1/x^2=14.

So x 2--3+1 x 2=11

x^4--3x^2+1)/x^2=11

x^2/(x^4--3x^2+1)=1/11.

1/a+1/b=(b+a)/(ab)=3

From the known obtained, a+b=5ab, then the original formula = (2*5ab+ab) (5ab-2ab)=11 3

The known equation formula, verify that x is not equal to 0, and then divide the two sides of the equation by x, the value of x+1 x can be obtained, and the value of the first formula can be obtained after squarering, and then use the result to put the second numerator and denominator squared by x at the same time, and bring it into the calculation, and the result can be obtained.

2.From 1 a+1 b=5, (a+b) ab=5, i.e., a+b=5ab, substituted for (2a+ab+2b) (a-2ab+b)=11ab 3ab=11 3

3.(1) For x -4x+1=0, verify x≠0, so divide both sides by x, x-4+1 x=0, that is, x+1 x=4, both sides are squared at the same time, there are x + 1x +2 = 16, x + 1x = 14

2) The numerator and denominator are divided by x at the same time, and 1 (x -3+1 x) = 1 11 is concluded using equation (1).

4a +4b = 4 (a + b) mega closure.

a²+b²=（a+b）²-2ab

4a²+4b²=4*29=116

a+b）²=a²+2ab+b²

a²+b²=（a-b）²+2ab

Ethnic stupidity a+b,2,2(a+b) =a-b) +4ab=81,1,1, we can find a=5, b=2, or a=2, b=5

Since ab is symmetrical, and the equation is also an early symmetric state, the solution must be the only solution as 116

2. The style of this question is the same as that of the first question.

a=7, b=2 or a=2, b=7

The solution is 81,1,4a +4b =4(a +b) = 4[(a+b) 2-2ab].

a+b）²=a-b)^2+4ab

Just substitute it. ,0,1, known a+b=7,ab=10, find the value of 4a +4b?

2. Knowing a-b=5 and ab=14, then the value of (a+b) is equal to ?

a^3+b^3=(a+b)^3-3ab(a+b)

The rest is solved by your substitution.

Formulas must be memorized proficiently.

From (a b 1) (posture a b a 1) trace to take the lack of thick 63, into (a b) 2 a 1 2 63, a b) 2 63 1 64, a b) 8, so there is:

a b a 8, or a b 8.

Solution: a+b=1, ab=-3

a²b+ab²=ab（a+b）=1×（-3）=-32a³b+2ab³=2ab（a²+b²）=2ab[(a+b)²-2ab]=2×(-3)×[1²-2×(-3)]=6)×7=-42

Have fun. a+b=3 both sides are squared at the same time.

a²+2ab+b²=9

ab=-2a²+b²=13

a³+a²b+ab²+b³

a²(a+b)+b²(a+b)

a²+b²)(a+b)

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a²b+ab²

ab(a+b)

a+b＝2,ab＝2.

1/2a³b+a²b²+1/2ab³

1/2a²(ab)+(ab)²+1/2(ab)b²＝a²+b²+4

a+b)²-2ab+4

1/2a³b+a²b²+1/2ab³

1/2ab（a²+2ab+b²）

1/2ab（a+b）²

Untie. 2a³b+2ab³

2ab(a²+b²)

2ab[(a+b)²-2ab]

1/2a³b+a²+b²

1/2a²*2+a²+b²

a²+a²+b²

a²+a²+2ab+b²-2ab

a²+(a+b)²-2*2

a +2 -4 = a solution;

a³b-2a²b²+ab³

ab(a²-2ab+b²）

2*(a-b)²

2[(a+b)²-4ab]

Solution: The algebraic simplification of the obtained algebra yields:

a 2(a+b)-2b 2(a+b) (1) give a+b=0 generations (1).

So the value of the algebraic formula is 0

Original formula = 2 (1 + 2a b-3a )-3 (2a -2ab -2) = 2 + 4a b-6a -6a +6ab +6 = 4a b-12a +6ab +8

Known set a {1,2,3} sock block, set b {2,4,6}. Sue Hu.

A stares at B {2};

aub＝｛1，2，3，4，6｝。

(a+b) squared: a square plus b square plus two times ab is equal to 1, (a-b) square is equal to (a + b) square minus four times ab, because ab is -1, so 4ab is -4, 1-(-4) = 5 3 (a+b) square = 3*5 = 15

Seek satisfaction!

=（a²+2a+1-2a)(b²+2b+1-2b)=(a²+1)(b²+1)

ab)²+a²+b²)+1

a+b=3（a+b)²=9

a²+2ab+b²=9

again ab=-4

a²+b²=17

Substituting ab=-4,a +b =17 into the original formula gives the original formula =(-4) +17+1=34

jly041218 did the right result, but it can't be changed like this in the middle. ）