If a 1 b 2 0, find the value of 3a 2b

000000161, hello:

If |a-1|+|b-2|=0, find the value of 3a-2b.

a-1|+|b-2|=0

a-1|=0,a=1

b-2|=0,b=2

3a-2b=3*1-2*2=3-4=-1

If a-1 and -6 are inverses to each other, find the value of a.

a-1=-(-6)

a=6+1a=7

If |x|=7，|y|=3, and x y, find the values of x and y.

x＜yx=-7

y=±3

First, a-1 is greater than or equal to 0, b-2 is greater than or equal to 0, and the sum of the two is 0, so a=1, b=2, 3a-2b=-1

The second way, because they are opposite numbers, so (a-1)+(6)=0, a=7 The third way, x is equal to plus or minus 7, and y is equal to plus or minus 3, so the answer has 2 solutions, x=-7, y=3x=-7, y=-3

1.If |a-1|+|b-2|=0, find the value of 3a-2b.

a=1 b=2

2 If a-1 and -6 are opposite to each other, find the value of a.

a-1=6a=7

3.If |x|=7，|y|=3, and x y, find the values of x and y.

x = plus or minus 7 y = plus or minus 3

x=-7 y=3

x=-7 y=-3

Obtained from the question|a-1|Greater than or equal to 0, |b-2|Greater than or equal to 0, so a-1=0, b-2=0, b=2 a=1, 3a-2b=3-4=-1

Take advantage of Veda's law.

ab=-1 Naixian1=-1 a+b=-(3) Guess 1=3

ab 2 + a 2b = ab (a + b) Chang Zhao Pai = -3

Solve the equation: a-1 2| +b-1/3)^2 = 0

Due to |a-1/2|The value of Qi Peiling is always non-negative, so we can consider its two possible values separately.

When a >=1 2, there is a-1 2 >=0, hence |a-1/2|=a-1 2, we get the equation:

a-1/2 + b-1/3)^2 = 0

Square sides: a-1 2) 2 + b-1 3) 2 = 0

Solve: a = 1 2, b = 1 3

When a >=1 2, there is a-1 2 >=0, hence |a-1/2|=1 2 - a, we get the equation:

1/2 - a + b-1/3)^2 = 0

Square sides: 1 2 - a) 2 + b-1 3) 2 = 0

Solve: a = 1 2, b = 1 3

Therefore, whether a is greater than or less than 1 2, there is a = 1 2 and b = 1 3 is a feasible solution.

Now, we can use this solution to brighten up and compute the value of 5(3a 2b)-(ab 2+3a 2b).

a = 1/2, b = 1/3

5(3a^2b) -ab^2 + 3a^2b) =5(3(1/2)^2(1/3)) 1/2)(1/3)^2 + 3(1/2)^2(1/3))

Therefore, the value of 5(3a 2b)-(ab 2+3a 2b) is 4 24 = 1 6. Hope.

The root number (a 2-3a+1) + (b-1) 2=0 is the sum of two non-negative numbers equal to 0, so both non-negative numbers are 0, so a 2-3a+1=0 and b-1=0

It is obtained by a 2-3a + 1 = 0 and dividing both sides by a.

a-3+1 a=0, i.e. a+1 a=3

The square of both sides gets: a 2+2+1 a 2=9

So a 2 + 1 a 2 = 7

From b-1=0, b=1

So a 2+1 a 2-|b|=7-1=8.

The root number (a 2-3a+1) + (b-1) 2=0 is greater than or equal to 0, and the addition is equal to 0, and both formulas must be 0, that is, a 2-3a+1=0

b^2-2b+1=0

If you have to substitute the following formula, you will have the answer.

From (a 2-3a+1)+b 2-2b+1=0

Result: (a 2-3a+1)=0 b 2-2b+1=0

Derive: Just find a and b respectively.

a+b=0 a-2=0

So a=2 is substituted into the original form.

b=-2, so the absolute value is 3a-2b=10

Solution: 丨2a-1丨+丨Difference3b-2丨=0

So 2a-1=0 a=1 2 3b-2=0 b=2 sentence town 3

a+b= 1 2+2 Void 3=7 6

It's an absolute value, right?

If so, the absolute value = 0, and the sum of the two absolute values = 0, which means that the absolute value of the two ants is equal to zero.

So 2a-1=0, 3b-2=0

The solution is a=1 2, b=2 and 3

The abbreviated defect is a + b = (1 2) + (2 3) = 7 6

The value is 0, and after this equation is raised to the common factor a, we get a 2+3ab+2b 2Multiply the crosses to get the original formula a(a+b)(a+2b) so it is 0

Bringing a=-2b into the algebraic equation gives the result of 0