999×999+1999 =(1000-1)²+1999 =1000000-2*1000+1+1999 =1000000-2000+2000 =1000000 =1000000-2*1000+1+1999 Is it unnecessary to subtract "1"?

999×999+1999 =(1000-1)²+1999 =1000000-2*1000+1+1999 =1000000-2000+2000 =1000000 =1000000-2*1000+1+1999 Is it unnecessary to subtract "1"?

1 in (1000-1) & sup2; perfect square formula 1000000-2x1000 + 1 is equivalent to B in (a-b) & sup2; = = A & sup2; - 2Ab + B & sup2

19 + 199 + 1999 +... + 199... 99 (100 9S),

Because 19 + 1 = 20199 + 1 = 2001999 + 1 = 2000
So the original formula is 20 + 200 + 2000 + 20000 +. + 200.00 (100 zeros) - 1 * 100
=2222.. 222 (100 2) 0-100
=22.. 22 (97 2) 120

Calculation: 199.8 + 19.98 + 1.998 + 2.222

199.8+19.98+1.998+2.222
=199.8+19.98+1.998+（2+0.2+0.02+0.002）
=（199.8+0.2）+（19.98+0.02)+(1.998+0.002)+2
=224

Calculation: 1998 + 199.8 + 19.98 + 1.998

1998 + 199.8 + 19.98 + 1.998 = (2000-2) + (200-0.2) + (20-0.02) + (2-0.002), = 2222-2.222, = 2222 - (10-7.778), = 2222-10 + 7.778, = 2219.778

A simple method of dividing 2240 by 70

2240/70
=(2100+140)/7
=2100/7+140/7
=300+20
=320

7.28 divided by 0.8 and then divided by 12.5?

7.28÷0.8÷12.5=7.28÷(0.8X12.5)=7.28÷10=0.728

(1 / 7 + 4 / 37) * a simple method of dividing 7-28 by 37

(1 / 7 + 4 / 37) × 7-28 △ 37
=1 × 7 of 7 + 4 × 7 of 37 - 28 of 37
=1 + 28 / 37-28 / 37
=1+0
=1

(2 / 3 minus 1 / 2) divided by (5 / 6 x 3 / 5)

Hello
(2 / 3 minus 1 / 2) divided by (5 / 6 x 3 / 5)
=1/6÷1/2
=1/3
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0.325 divided by 0.013 / (0.22-0.2065) divided by (3.6 * 0.015)
Simple algorithm (/ is a fractional line)

0.325 divided by 0.013 / (0.22-0.2065) divided by (3.6 * 0.015)
=325 divided by 13 / (0.0135) divided by (3.6 * 0.015)
=25×0.0135÷（3.6×0.015）
=25×135÷（36×15）
=25×135÷15÷36
=25×9÷36
=25/4

The product of 0.0325 divided by 0.22 of 0.013 minus 0.2065 divided by 3.6 times 0.015 is equal to

-7.8039772727272727272727272727273e-4