Help me do these questions, ask for the process, and the reward will be increased

4.If the constant point (1,10) satisfies the function f(x), and substituting it into the function gives 10=a (1 2+2-3)+m=a 0+m=1+m

This gives us m=95Power of zero = 1 for any non-zero number

That is, y=a (2x-2) is constantly passing through the point a(1,,1) and the straight line l also passes through the point a

then m+n=1

The distance from the coordinate origin (0,0) to the straight line = 1 root (m2+n 2) when m = n = 1 2, the maximum value is root 2

6.(3 2) x=(2+3a) (5-a) has a negative root.

Because 3 2 is greater than 1

So (3 2) x is a monotonically increasing function.

When x<0 there is (3 2) x<1

then 2+3a 5-a<1

a<3/4

and f(a+2)=18

So f(a+2)=3 (a+2)=3 ax3 2=18a=log3 2

g(x)=into x3(ax)-4 x defines the domain as [0,1] and is a monotonically decreasing function.

then g(0)>g(1).

That is, g(0) = into -1> g(1) = into 3 (log3 2)-4 1=2 into -4

-1>2 -4

Into "-3 into <3

2、f(x)=x+1/x x>0

e^x+x x<=0

X+1 x>=2 at x>0 excludes AD

When x=0, f(0)=1 so exclude b and choose c

3、f(1)=a^2=1/9 a=1/3

f(x)=1/3^|2x-4|=(1/3)^(2x-4) x>=2 (1/3)^(4-2x) x<2

And (1 3) (2x-4) decreases so the decreasing interval is [2, infinite) choose b

4. Obtained by substitution (1,10).

a^0+m=10 m=9

5. y=a (2x-2) constant point a (1,1), the straight line l mx+ny-1=0 over (1,1) is substituted to obtain m+n=1

o The straight-line distance is d=1 m2+n2

and by m 2 + n 2 > = 1 2 (m + n) 2 = 1 2

0 is obtained, at which point m=m=1 2

The maximum value is 2

6、t=(3/2)^x

t=(2+3a)/(5-a) a=(5t-2)/(t+3)=5-17/(t+3)

If there are negative roots, then the range of x<0 0a is (-2 3, 3 4).

f(a+2)=3^(a+2)=18 a+2=log3(18)=2+log3(2) a=log3(2)

2)g(x)=λ3^(ax)-4^x=λ3^(xlog3(2))-4^x=λ2^x-4^x

g(x) monotonically decreases, then g'(x)= ln2*2 x-2ln2*4 x<=0 is constant at 0<=x<=1.

That is, <=2*4 x 2 x=2 (x+1) is constant.

0<=x<=1 then 2<=2 (x+1)<=4 so <=2

The range is (-infinity, 2].

<> as shown in the mega pulse diagram in Oak Guess. Yangsan.

Can you shoot ** clearly? I'm a little vague here.

The first figure: A** cost = 1200 (1 + 20%) = 1000 yuan B ** cost = 1200 (1-20%) = 1500 yuan total profit and loss = 1200 * 2-1000-15000 = -100 yuan total loss of 100 yuan.

Clause. Two and three pictures:

Original price (1+20 )=1200 The original price is 1000 and 200 is earned

The original price (1-20) = 1200 The original price is 1500 and the loss is 300, so the two are combined and the loss is lost.