Grade 8 Mathematics Geometry Problems, Grade 8 Mathematics Geography Problems

2cd;2OC=1, so ohm=ogm=60°, by hm=1, so me=mf( ).

Because oh = oe, h, eh

then there is EHF=60°; 2ob=1, i.e., FH=eg( ) and EHF= egm=60°( EHF

mge：δehm≌

mgf;2cd, hm=fg( )

by de, connect fg; 2ab,of=og,so ef=me( )by,gm,so by; 2AB, so oh+of=OE=OG, MH, so EHM= FGM=120°(

and gm=1 and hob=120°, so gm=eh( ) is the same, eh=1 and fg=1 take the midpoint g of ob and oc respectively

1.It is known that the polynomial x 4+2x 3-x+m can be factored, and there is a factor of x-1

1) When x=1, find the value of the polynomial x 4+2x 3-x+m;

2) According to the result of (1), find the value of m;

3) Follow the method in (1) to determine whether x+2 is a factor of the polynomial x 4+2x 3-x+m.

2.Known(2005-a)(2002-a)=2003, find the value of (2005-a) +2002-a).

3.There are two kinds of fruits in the wholesale market: A type is M kg per box, and the price is a yuan; Each box of B weighs n kilograms, the price is b yuan, try to find the relationship between a single and a b.

Proof: af=ad

ad=ab, so af=ab;

afe=∠ade=90°=∠abg;

ag=agSo: abg afg.

Let BG=X, CD=3DE, DE=AB=6, then DE=FE=2.

abg afg, then bg=fg=x, cg=bc-bg=6-xcg 2+ce 2=eg 2 (Pythagorean theorem).

Then: (6-x) 2+4 2=(2+x) 2, and the solution is x=3then BG=3, GC=6-X=3

then BG=GC. (The method is stupid).

bg=gf=gc=3

then GCF= GFC, GCF+ GFC+ FGC=180°=2 FCG+ FGC

agb=∠agf,∠agb+∠agf+∠fgc=180°=2∠agb+∠fgc

then fcg= agb

then ag cf

gf/ge=gf/(gf+fe)=3/5

Then the area of fgc is 3 5

The area of the EGC is gc*ce 2=3*4 2=6

Then the area of fgc is .

The first question, you fill in the diagram first.

The second one is 60 degrees of acuteness because the obtuse angle is 120 degrees.

So the triangle formed by the short side is an equilateral triangle.

So half of the diagonal is equal to.

So the diagonal is equal to centimeters.

Double the length of the midline of the BC side to H, so FG=FH, because EG is perpendicular to AB so EF=FH=FG, because the angle EBC=70 degrees, E, F are AB, BC, so EB=BF, so the angle FEH=angle FHE=55 degrees, because the angle CGH=Angle EHF, so the angle CGF=55 degrees55 degrees.

After D as dg ac to bc and g, dbfg efc. can be obtained according to the conditionsThat is, df=ef

First draw a diagram to verify that the two triangles divided by the line are congruent (the two angles divided by the bisector line are equal, the sides divided by the middle line are equal, plus the common side), so the other two sides of the two small triangles are equal, and the conclusion is that the triangle is an isosceles triangle.