toandhsp

> restart:

with(geom3d):

point(A,1, 0, 0):

point(B,0, 1, 0):

point(C,0,0,1):

plane(P, x+y+z=6,[x,y,z]):

point(T,x,y,z):

sol:=solve([distance(T,A)=distance(T,B),

distance(T,A)=distance(T,C),Equation(P)],[x,y,z]);point(T1,rhs(op(1,sol[1])),rhs(op(2,sol[1])),rhs(op(3,sol[1])) ):

restart;

L:=[]:

k:=10:

for x1 from -k to k  do

for y1 from -k to k  do

for x2 from -k to k  do

for y2 from -k to k  do

for x3 from -k to k  do

for y3 from -k to k do

G:=[1,1]:

H:=[3,3]:

Thanks for all helps.

I tried solve it with Mathematica

Reduce[Sqrt[x + 3 - 4*Sqrt[x - 1]] + Sqrt[x + 8 - 6*Sqrt[x - 1]] == 
 1, x, Reals]

First code

>restart:

f:=x->a/x^2-x+2:

g:=x->a*sin(2*x):

ans:=allvalues(solve(f(f(-1))=sqrt(2)+g(f(2)),a)):

ans2:=[evalf(ans)];

We get

ans2 := [-2.881759230-1.051335058*I, -.7437844448-1.866833976*I, -2.881759230+1.051335058*I, -.7437844448+1.866833976*I, -5.471111560]

Second code

>restart:

Thank you very much.

Thank you.

Thank you very much.

Thank you very much.

Put the equation of the line passing the point A(4,1) has the form a*(x-4)+b*(y-1)=0. We always assume a^2 + b^2 = 1. And then, you solve the system of equations

solve([-4*a-b=2,a^2+b^2=1],[a,b]);

But I don't know how to get the exactly solutions [a = -8/17-1/17*13^(1/2),b = -2/17+4/17*13^(1/2)] and [[a = -8/17+1/17*13^(1/2), b = -2/17-4/17*13^(1/2)]]

I must solve in two cases 

How about this code? Maple can not run. Please help me. 

> restart:

a:=2:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(30*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(120*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

> restart:

a:=3:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(15*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(135*Pi/180)], [b,c]);

with(geom3d):

point(o,0,0,0):

point(A,a,0,0):

Thank you very much.

When I  found the length of sides b and c, I created a triangle satisfies the conditions stated and test the measure of the  angles of this triangle. I tried

> restart:

a:=3:

eq1:=(a^2+b^2-c^2)/(2*a*b):

eq2:=(b^2+c^2-a^2)/(2*b*c):

eq3:=(c^2+a^2-b^2)/(2*a*c):

sol:=solve([eq1=convert(cos(15*Pi/180), radical), eq2=cos(30*Pi/180), eq3=cos(135*Pi/180)...

Thank you very much.