Eight math problems

The answer is D (I'm pretty sure oh, we did).

a^4-b^4=a²c²-b²c²

a²-b²）（a²+b²）=（a²-b²）c²a²+b²=c²

From the inverse theorem of the Pythagorean theorem, it can be judged that ABC is a right triangle.

a^4-b^4=a²c²-b²c²

a^4-a²c²=b^4-b²c²

a²(a²-c²)=b²(b²-c²)

a=b, so abc is an isosceles triangle.

So ABC is an isosceles or right-angled triangle.

Hope I helped you! 0（* #

A to the fourth power - B to the fourth power is (A + B) * (A - B) = (A - B) * C is simplified to be A + B = C, of course, a right triangle.

Oh, it can't be isosceles, or square a-b will be equal to 0.

Answer: da 4-b 4=a c -b c

a²-b²）（a²+b²）=（a²-b²）c²a²+b²=c²

Pythagorean theorem At the same time, if a=b then it is an isosceles triangle.

A to the fourth power - b to the fourth power = a c -b c , a + b ) (a -b ) = (a -b ) c , a + b = c From the inverse theorem of the Pythagorean theorem, it can be judged that abc is a right triangle.

a to the fourth power - b to the fourth power =< a²-b²>

a²c²-b²c²=.c²

So a +b =c or a =b i.e. a = b so isosceles or right triangle.

(2 + root number three) to the power of 2019 (3-root number) to the power of 2020 The answer is 2-root number three.

1 Jihuaitan Botong 2a-ab+b*b+2=0 ; 1/2a+2=ab-b*b ; 1/2a+4/2=(a-b)b ;1 2(a+4)=(a-b)b will not be sorry.

1）(2x+3y)(2x-3y)=31=1×31.

Since x and y are both positive integers, 2x+3y>2x-3y, so, 2x+3y=31, 2x-3y=1, x=8, y=5.

2) The radius of the great circle is 13, the radius of the small circle is 3, and the area of the circle = 169 -9 = 160.

Because x and y are both positive integers.

Original = (2x-3y)(2x+3y)=31

2x+3y=31 2x-3y=1

x=8 y=5

Do it yourself below.

Proof: Extend be and ac, at point m

be ae, then aeb= aem=90°;

and bae= mae, ae=ae, then bae δmae(asa), eb=em

be⊥ae;CF AE, then be fc

af/ae=fc/em=fc/eb=pc/pe.

i.e. af ae = pc pe, af fe = pc ce, then cf ap

The picture is not good, I hope it helps you.

x+y-2xy)(x+y-2)+(xy-1)²

Don't be afraid to be annoyed, if you solve it all, the original formula becomes: (the method used is the principal element method, which regards x as the principal element and y as a constant).

x+y-2xy)(x+y-2)+(xy-1)^2=[（x+y)-2xy][(x+y)-2]+(xy-1)^2=(x+y)^2-2(xy+1)(x+y)+4xy+x^2y^2-2xy+1

x+y)^2-2(xy+1)(x+y)+x^2y^2+2xy+1=(x+y)^2-2(xy+1)(x+y)+(xy+1)^2=[(x+y)-(xy+1)]^2

x+y-xy-1)^2

x(1-y)-(1-y)]^2

x-1)^2(1-y)^2

x-1)^2(y-1)^2

The conclusion is valid, proving: take a little e on the ray AN such that ae = acConnect CE because the angle BAD=120 degrees and AC is the angle bisector of that angle, so the angle CAD=angleACE=angleAEC=60

degree, so ac=ae=ce, which is known to have the angle cba=anglecde, and the angle, cab=aec, so the triangle cba, all of which are in the triangle cde, so ab=de, because ac=ae=ad+de=ad+ab