Math(Input=Result) Error!
binom((3x2−4×x3)10)=binom(Error: loop)
(3x2−4x3)10
I think the answer is:
10Cr×(3x2)10−r×(−4x3)r=10Cr×310−r×(−4)r×x−2(10−r)×x3r=10Cr×310−r×(−4)r×x−2(10−r)+3r=10Cr×310−r×(−4)r×x−20+2r+3r=10Cr×310−r×(−4)r×x−20+5r
so -20+5r=0 -> r = 4
the term we want is
10C4×36×(−4)4×x0
Using sigma notation and factorials for the combinatorial numbers, here is the binomial theorem:
(1) -8x - 12y = 36 -> 8x + 12y = -36
(2) 5y - 6x = 41 -> -6x + 5y = 41
8x+12y=−36−6x+5y=41
\boxed{8x+12y=−36−6x+5y=41}
x=|−3612415||812−65|=−36∗5−41∗128∗5−(−6)∗12=−180−49240+72=−672112=−6x=−6y=|8−36−641||812−65|=8∗41−(−6)(−36)8∗5−(−6)∗12=328−21640+72=112112=1y=1
\\x=
\frac{|−3612415|}{|812−65|}
=\frac{-36*5-41*12}{8*5-(-6)*12}
=\frac{-180-492}{40+72}
=\frac{-672}{112}=-6\\
\boxed{x=-6}
\\y=
\frac{|8−36−641|}{|812−65|}
=\frac{8*41-(-6)(-36)}{8*5-(-6)*12}
=\frac{328-216}{40+72}
=\frac{112}{112}=1\\
\boxed{y=1}