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Known -1a-b>2....4)

Anisotropic inequalities can be subtracted, and the direction of the unequal sign after subtraction is the same as the direction of the inequality sign of the subtracted formula, therefore:

1)-(4)-5<2b<1....5)

Divide 2 by (5) on both sides to get :

5 25) + (6) -15 2<3b<3 2....7)3)+(7) gives -13 2<2a+3b<17 2, which is the value range of 2a+3b.

Codirectional inequalities cannot be subtracted.

2a+3b=x(a+b)+y(a-b)

x+y)a+(x-y)b

The solution yields x=5 2, y=-1 2

2a+3b=5/2(a+b)-1/2(a-b)5/2<5/2(a+b)<15/2

2<-1/2(a-b)<-1

Therefore -9 2<2a+3b<13 2

Solution: Treat m as a variable and x as its parameter.

Then the inequality is transformed into: (x-1)m+x 2-4x+3>0 by the title, and this equation holds for 0 m 4 constantly.

That is, when f(m)=(x-1)m+x 2-4x+3 is greater than 0f(m) when it is 0 m 4, the minimum value is a one-time function, and the minimum value is obtained at both ends, that is, f(0)>0 and f(4)>0

i.e.: x 2-4x+3>0 and x 2-1>0

The first inequality is solved: x<1 or x>3, the second inequality is solved by x<-1 or x>1 two solution sets intersected, and x<-1 or x>3 is obtained, which is the final result is satisfactory, please adopt, I wish you progress in learning!!

Hope it helps you and can adopt. It's also not easy to type these words in word.

f(x) is defined in the domain x>0

f'(x)=1 x-a x 2=(x-a) x 2 so f(x) decreases at (0,a) and increases at (a, positive infinity) by knowing that 1 x0-a x0 2<=1 2 pairs x0>0 constant a>=-1 2x0 2+x0, for x0>0 constant holds, and y=-1 2x 2+x is a parabola with an opening downward, the axis of symmetry is x=1, and the maximum value is 1 2 at x>0

So a>=1 2, so the minimum value of a is 1 2

Let g(x)=f(x)-(x 3+2(bx+a)) 2x+1 2=lnx-1 2x 2-b-1 2,x>0

g'(x)=1 x-x=(1-x)(1+x) xSo g(x) increases at (0,1) and decreases at (1,positive infinity) g(1)=-b-1

So when b<-1, the equation has two real roots, and when b=-1, the equation has one real root.

When b>-1, the equation has no real root.

By Newton's second law of motion:

f = ma and v square = 2as

It can be seen that v square = 2sf m

Move the item to get 1 2mv square = fs

Because the initial kinetic energy is equal and s is equal, so f is equal.

s is the same, f*s = the initial kinetic energy is the same, and the last is the same 0So there is a resultant force like **choose c

c Because the kinetic energy of the object is the same, the same force changes the kinetic energy of the object in the same displacement.

Sorry, I was in middle school, I just did the task.

1.Because it is not obvious to look directly, then try subtracting the two formulas. Left-right=a 4-a 3xb+b 4-ab 3=a (a-b)+b (b-a)=(a-b) (a-b)=(a +b +ab)(a-b) 丨2ab丨+ab)(a-b) 0 If and only if a=b is true, a -b = (a-b) (a +b +ab) is used in the middle, because we don't know a, b is positive or negative, so the mean inequality is used to get 丨2ab丨 Evergrande is ab, so it is judged that the left is greater than or equal to the right.

2.Let f(x)=x-1 x, then f'(x)=1+1 x 0 so f(x) increases singliably in the defined domain, so when a b and on the same side of the y-axis, f(a) f(b) i.e., a-1 b-1 b is constant. ab 0 or ab 0 can be known

3.The first: 2+3 pushes out 1, because ab 0, so divide the two sides of 2 by ab the second one:

1+3 pushes out 2, because ab 0, so multiply both sides of 1 by ab Third: 1+2 pushes out 3, the counterproof method, let ab 0, multiply ab on both sides of 1 and contradict 2, so it is proven.

>=a^3*b+a*b^3.

b, a-1 a>b-1 b, a-1 a-(b-1 b)=(a-b)+(a-b) (ab)=(a-b)(ab-1) (ab)>0,(ab-1) (ab)>0,a*b should meet the condition of ab>1 or ab<0

3.(1)c／a＞d／b<==>(bc-ad)/(ab)>0.

bc＞ad⑶ab＞0

propositions 1 (2), (3) ==>(1);

proposition 2 (1), (3) ==>(2);

Proposition 3 (1), (2) ==> (3).

It can be seen from the meaning of the title, because y=|x+a|The images are all on the positive semi-axis of the y circumference, and only one intersection point with the x-axis is x=-a, and because the two images have three different intersection points, and the y=4 x image is only in the 1,3 quadrant, so the 3 intersections are all in the first quadrant, and there are two common points on the left side of the first quadrant-a, and there is one on the right. The hyperbola and the left branch of the straight line are: -x 2-ax-4=0, and the discriminant formula is:

a 2-16, another discriminant is equal to zero, a 2 = 16 so (-a) = 4, (because y = |.)x+a|, so is -a) because y=|x+a|The axis of symmetry is in the first quadrant, so -a>0, so a<0, so a=-4, is just critical, when a=-4, there are exactly two intersection points, so when the symmetry axis is translated to the right side of 4, it can be guaranteed that the intersection points are three, if it moves to the left, then y=|x+a|The left branch has no common point with y=4 x, and the above a<-4

Look at the image to understand:

y=|x+a|It is to symmetrically go up the part below the x-axis of y=x+a.

Therefore, it can only intersect with the first quadrant part of y=4 x.

Then y=|x+a|When (2,2) is y=x+a(2,-2), a=-4 is obtained, and b is selected with two intersections

No specific images are used.

The former is in one or three quadrants.

The latter is on and above the x-axis The slope is plus or minus 1 and there is no intersection below the x-axis, and it is roughly known that there are 1 or 2 or 3 intersections When tangent (i.e. a particular value, here 4) there is only one case - two intersections, so a and c are excluded

When the image of the latter is shifted to the left infinitely (i.e., a becomes larger), there is an intersection that excludes d d and result b

Solution: Orange is known to be b =ac x=(a+b) 2 y= b+c 2

a x+c y=2a can be a celery ball a+b +2c (b+c)=2 ab+2ac+bc (ab+2ac+bc)=2

Answer: Original sensitive chain = 2

solution, from the original formula, loga[a (2x)-2a x-2]<0loga[a (2x)-2a x-2]1

a^x+1)(a^x-3)>0

So: a x-3>0

then x