Give everyone a math problem, if you are interested, please come in!

How much can't**, can**?

Question 1: 88*2 (1+1+10%) = male.

The second problem solves the first equation.

3x-6+5=6x+2

3x=-3x= -1

So the second equation gives x=-1-2=-3

Substituting 3kx-1=k-x

9k-1=k+3

k=-2/5

Question 3 (A+2) 2+|a+b+5|=0

Both the perfect square and the absolute value are non-negative.

So (a+2) 2=0

a+b+5|=0

solution, a= -2, b= -3

3a^2b-[2a^2b-(2ab-a^2b)-4a^2]-ab=+2ab+4a^2-ab

22 Question 4.

Let the hundredth digit of the three-digit number be a, the ten-digit number is b, and the single-digit number is c, according to the title, the three-digit number is 100a+10b+c, add their three numbers together to get a number, and subtract the original three-digit number to obtain the number, the result is 100a+10b+c-a-b-c= 99a+9b=9(11a+b), so the result must be a multiple of 9, and any three-digit number is true.

Solution: Suppose this plant originally had x kilograms, according to the title:

2×[2×（2x＋3）＋3]＋3＝45

8x＋21＝45

x 3 A: This plant turned out to be 3 kg.

Hope it helps!

Let this plant have x kilograms, 2*(2*(2x+3))+3=45, and the solution is x=93

Suppose this plant turned out to be x kilograms.

The plant has 2x+3 kg on the first day.

On the second day, this plant has 2 (2x+3)+3=4x+9 kg, and on the third day, this plant has 2 (4x+9)+3=8x+21 kg, so 8x+21=45x=3

1 120 2 60 1 (hours).

2 It takes at least X hours for the speedboat to meet the research ship after departing from Island C.

2＋x）×20＝20x＋40, oc＝60√3（20x＋40）²＋60√3）²－2×（20x＋40）（60√3）cos30°

60x) Solve this equation to get: x1 1 x2 13, so it takes at least 1 hour to meet the research ship.

(1) It takes one hour for the speedboat to travel from port B to island C. (The algorithm is simple, think about it yourself, it will help you to answer the same problem next time you encounter it, and you can answer it empirically).

2) The second question is not clear. Did the speedboat go in the same direction after loading supplies on Island C? This is very important!

(1) When t=2, dh=2*2=4

AD=8 is known, so the point h is the midpoint of AD.

Known M AD, i.e. EF AD

So, EF is the perpendicular bisector of the line segment AD.

So, ae=de, af=df

Again, ABC is an isosceles triangle, EF BC

So, AEF is also an isosceles triangle.

i.e., ae=af

So, ae=de=df=af

So, the quadrilateral aedf is a diamond.

2) Let the motion time t, then: dh=2t

Ad=8 is known, so: ah=8-2t

EF BC is known, so: ah ad=ef bc==> (8-2t) 8=ef 10

> ef=(5/4)(8-2t)

So, s pef = (1 2)ef*dh

1/2)*(5/4)(8-2t)*2t=(5/2)t(4-t)=(5/2)(-t²+4t)=(-5/2)[(t²-4t+4)-4]

-5/2)(t-2)²+10

Therefore, when t=2, the area of PEF is the largest, and the maximum value is 10, at this time, BP=3 2=6cm

1.Since the remainder of f(x) divided by (x-1)(x-2)(x-3) is 2x 2+x-7, then f(x)=g(x)(x-1)(x-2)(x-3) +2x 2+x-7, obviously when f(x) is divided by (x-1), g(x)(x-1)(x-2)(x-3) divided by (x-1) will not have a remainder, so only 2x 2+x-7 divided by (x-1) can be used, and 2x 2+x-7=(x-1)(2x+3) - 4, that is, the remainder of 2x 2+x-7 divided by (x-1) is -4, so the remainder of f(x) divided by (x-1) is -4 if f(x) divided by (x-1) (x-2) (x-3) the remainder is 2x +x-7

Because 2x +x-7 divided by x-1 is -4

So f(x) divided by x-1 is -4

Because 2x + x-7 divided by (x-2) (x-3) the remainder is 11x-19

So f(x) divided by (x-2)(x-3) remainder is.

lgx)^2-(lg3+lg12)lgx+lg3lg12

lgx)^2-2lg6lgx+lg3lg12

lgx-lg6)^2+lg3lg12-(lg6)^2

lgx-lg6)^2-(lg2)^2

Therefore, when x=6 there is a minimum value of -(lg2).

log3(10)=log3(2*5)=log3(2)+log3(5)=a ......1）

log6(25)=log3(25)/log3(6)=2log3(5)/[log3(2)+1]=b……（2）

From (1) and (2), log3(5) = (ab+b) (b+2), log3(2) = (2a-b) (b+2).

Substitute log4(45)=[log3(5)+2] 2log3(2) to get it.

log4(45)=(ab+3b+4)/2(2a-b)

First, the function takes the minimum value of 1 at x=1

So find the solution where f(m)=m is greater than 1.

So m=3