I2x + 14i = 6 i9x-18i = 27 i8x +12i = 20 i15 x - - вопрос №4948505214691827812201510591568 от 1lёn1 26.05.2021 13:07

Question

Математика: I2x + 14i = 6 i9x-18i = 27 i8x +12i = 20 i15 x - 10i = 5 i9x + 15i = 6 i8x - 6i = 14 iixi - 3i = 0 i5- ixii = 0 iixi + 3i = 0 i7 - ixii=0

1lёn1 · Answer

I2x + 14I = 62х=-14+62х=-8х=-8/2х=-4I9x-18I = 279х=27+189х=45х=45/9х= 5I8x +12I = 20 8х=20-128х=8х=8/8х=1I15 x - 10I = 5 15х=10+5х=15/15х=1I9x + 15I = 69х=6-159х=-9х=-9/9х=-1I8x - 6I = 148х=14+68х=20х=20/8х=2.5IIxI - 3I = 0х=3I5- IxII = 0х= 5IIxI + 3I = 0х=-3I7 - IxII=0х=7...

I2x + 14i = 6 i9x-18i = 27 i8x +12i = 20 i15 x - 10i = 5 i9x + 15i = 6 i8x - 6i = 14' iixi - 3i = 0 i5- ixii = 0 iixi + 3i = 0 i7 - ixii=0