One to the math problem

1) According to the question: x belongs to [0, + infinity).

x>(4k-2) 3

To be satisfied, then (4k-2) 3 cannot be at [0, +infinite) i.e., (4k-2) 3 <0

4k-2<0

k<1/2

The range of k-values can be calculated.

2）-3x<-2-4k

3x>2+4k

x>(2+4k)/3

To make x not negative, i.e. x>=0

2+4k)>=0

4k>=-2

k>=-1/2

To do this kind of problem, you can first think of (2+4k) 3 as a whole, and x > as the whole.

and x>=0

then the whole must be "= 0".

If you do more, you will be familiar with it.

Good luck with your studies!

So. 4k-2)/3=0

So. 4k-2=0

k=1 2 Hope it helps you.

Learning progress

x>（4k-2）/3

3x+2)/4>k

That is, k<(3x+2) 4<(3*0+2) 4=1 2 is k<1 2

Go see the answer. Or ask the teacher.

That is, (4k-2) 3 is less than x x+6 12k k x+6 12 and there is no exact value.

When x=0, f(0)=1x> f(x)=(x2+kx+1) (x2+x+1) =1 (k-1)x (x2+x+1) =1+ (k-1) (x+1 x +1) 1+ (k-1) 3=(k+2) 3 (x=1 x, i.e. when x 1, the equal sign holds) Therefore, the maximum value of f(x) f(1) (k+2) 3 When x approaches infinity, The value of f(x) tends to be close to 1 So, the minimum value of f(x) 1 is solved by the problem, 1 2>(k+2) 3, k<4 So, the range of the value of k is k<4

k k greater than is also true, to take the greater than sign.

The length of the two sides is b c and happens to be the two real roots of this equation.

Therefore, Brother Morisun b+c=2k+1, bc=4(k-1 2) is an isosceles triangle and needs to be solved by classification.

1.If b=c, then b=c=k+1 2

bc=(k+1 2) 2=4(k-1 2) to get k=2 3 (this equation is easy to solve) b=c=k+1 2=2

Circumference = a+2+2=a+4

2.If b=a, c=2k+1-b=2k+1-abc=a(2k+1-a)=4(k-1 2).

a-2)(2k-a-1)=0 cannot be reduced to (a-2).

If a=2, k can be taken at any value.

Circumference = a + 2k + 1

If a is not equal to 2, k = (a+1) 2

Circumference = a + a + (a + 1 + 1-a) = 2a + 2

Solution: Because the other two sides are long b c are exactly the muffled roots of the two real numbers of this equation.

The relationship between the root and the coefficient is known to be the tremor of the hand.

b+c=(2k+1)/2

So the circumference of the triangular biblic abc is.

a+(2k+1)/2

44÷(1-1/5)=55

A: The first root is 55 meters, and the second root is a meter.

First root (1-1 5)=44

The first root = 55 meters.

2nd root + 2nd root = meter.

The first root was originally 55 meters long, and the second root was originally 55 meters long.

The first one is 55 meters long.

The second is a long meter.

It's good to have a picture. My teacher taught me. The first step is to draw a picture. Particularly easy to understand.

According to the Pythagorean chord. The ab ratio is 13, which is 25 times according to the magnification (75 5 = 25). 13*25=325

ab=【√5^2+12^2）】*75/5）=325<>

Let the slope angle be , then tan =5 12

sinθ=5/13

lab=75/sinθ

lab=195m

That is, the distance traveled by the car from A to B is 195m

Let ab distance x, then there are 5:12 75:x, and the solution is x 180m

Satisfied. (1) Solution, if the cost of the moped is a yuan, the number of vehicles sold in the first quarter is b, and the average profit per vehicle in the second quarter is x.

20%ab=x%a(4/5)b

You can also directly set the equation as:

20%＝（4/5)x%

can be solved x=25

2) Set the first. 3. If the average percentage reduction in the fourth quarter is y, the selling price in the fourth quarter will be (1+25%)a(1-y)(1-y).

The column equation is: 20%ab=[(1+25%)a(1-y)(1-y)-a]16b

You can also directly set the equation as:

20% = [1+25%)(1-y)(1-y)-1]16 solution yields y 10% or y 110% (rounded).

Because a+b+c+d=0, a+b+c=-d

The cube on both sides can be moved to get filial piety:

a^3+b^3+c^3+d^3+3ab(a+b）+3ac(a+c)+3bc(b+c)+6abc=0

And because a+b=-c-d, a+c=-b-d, b+c=-a-d are substituted into the above formula: Ji Shen Feng.

a^3+b^3+c^3+d^3-3（abc+bcd+cda+dab)=0

Because a 3+b 3+c 3+d 3=3

abc+bcd+cda+dab=1

I hope mine can help you!