1, "Breaking Ten Laws"
13 consists of 1 tens and 3 ones. You can subtract 9 from 10 first, and combine the remaining 1 with 3 in the unit to get the remaining 4. The basis of this algorithm is to master the composition of 1 1 ~ 20, and can calculate the addition and subtraction within 10, including the mixed operation of addition and subtraction.
2. "Ten-level method"
You can divide 13-9 by a previous continuous subtraction, and subtract 3 from 13, and then subtract 6, leaving 4. The basis of this algorithm is that students have mastered the division and combination of numbers within 10, and can calculate the subtraction within 10, the subtraction from ten to ten, and the continuous subtraction. ?
3. "Do subtraction and want to add"
Using the addition and subtraction relationship, as long as we know that 9 plus a few equals 13, we can deduce that 13 minus 9 equals several. The basis of this algorithm is that students will write the corresponding subtraction formula according to the addition formula, find out the unknown in brackets and calculate the carry addition within 20.
4. The formula of abdication subtraction within 20:
"Minus nine plus one, eight plus two, seven plus three, six plus four, five plus five." For example: 17-9 = (? ) take the 7th bit of 17 and add 1, and the result is 8, that is, 17-9 = 8, 13-9 = () take the 3rd bit of 13 and add 1, and the result is 4.
Extended data
1, benchmark method
When several numbers close to a certain number are added, you can take a certain number as the reference number, then multiply the reference number with the added number, and add the difference between each number and the reference number to get the calculation result.
For example: 8 1+85+82+78+79.
=80x5+( 1+5+2-2- 1)
=400+5
=405
2. Split method
Mainly, some disassembled scores cancel each other out to simplify the operation. Fractions with the general shape of 1/ax(a+ 1) can be divided into. 1/a- 1/a+ 1 .
For example:11x 2+1/2x3+1/3x4+1/4x5+1/5x6.
= 1_ 1/2+ 1/2- 1/3+ 1/3- 1/4+ 1/4- 1/5+ 1/5- 1/6
= 1- 1/6
=5/6