In △ABC, BC=a, AC=b, AB=c, and satisfies a4+b4+1 2C4=a2c2+b2c2. Try to determine the shape of △ABC.

In △ABC, BC=a, AC=b, AB=c, and satisfies a4+b4+1 2C4=a2c2+b2c2. Try to determine the shape of △ABC.

A4+b4+1
2C4=a2c2+b2c2 deformation is:
A4+b4+1
2C4-a2c2-b2c2=0,
(A4-a2c2+1)
4C4)+(b4-b2c2+1
4C2)=0,
(A2−1
2C2)2+(b2−1
2C2)2=0,
A=b,
A2+b2= c2,
So △ABC is an isosceles triangle.

The absolute value of the fourth power of -3+(-0.25)*100 power of 4+(one-half-one-third)/(one-sixth) second power /(-2) is calculated 3 To the fourth power +(-0.25) to the 100th power *4 to the 100th power +(1/2-1/3)/(1/6) to the second power /(-2) absolute value, calculated The absolute value of the fourth power of -3+(-0.25)*100 power of 4+(1/2-1/3)/(1/6) second power /(-2), calculated

-3 To the 4th power +(-0.25) to the 100th power *4 to the 100th power +(1/2-1/3)/(1/6) to the 2nd power /(-2) absolute value
=-81+(0.25×4) To the 100th power +1/6/(1/36)/2
=-81+1+3
=-77