First, the main basis for simplifying real numbers with root signs
1, (√a)=a(a≥0), (field) =a.
2,√a=∣a∣ field = a.
3、√ab=√a√b(a≥0，b≥0)
4、√a/b=√a/√b(a≥0，b & gt0)
The above formula can be used for simplification from left to right and from right to left. In addition, the multiplication rules and formulas of algebraic expressions are used.
Second, the requirements of simplifying the real number results with the root sign:
1, the root sign cannot contain a factor that can be used to find the root.
2. The root sign (root sign) does not contain the denominator.
3, there is no root sign on the denominator.
Third, the application examples
1, on the simplification of the internal factor of the root number
Example 1, simplified √48
Solution: √48=√4*4*3=√ 16*3=4√3.
Note: the numbers in the root number should be decomposed into qualitative factors, and everything that can be written should be included, such as: √48=√4* 12=2√ 12, which is not a complete simplification.
2. About removing the denominator in the root sign.
Example 2, √ 48-6 √ (1/3)+√ (1/27)
Solution: The original formula = √16 * 3-6 √ (3/3 * 3)+√ (1* 3/9 * 3 * 3).
=4√3-2√3+(√3)/9
=( 19/9)√3
Another solution: the original formula = √16 * 3-6 * (1√ 3)+1√ 27.
=4√3-6*√3/(√3*√3)+√3/(3√3*√3)
=4√3-2√3+√3/9
=( 19/9)/√3。
Here, the basic properties of fractions are applied, and the denominator that cannot be written is changed into a number that can be written, or the root sign on the denominator is changed. Note that √ (1/a) = √ a/a (a > 0) application.
3, about removing the root sign in the denominator:
Example 3, simplification (√ 12+√27)/√3.
Solution: The original formula =(2√3+3√3)/√3=5√3/√3=5.
Another solution: the original formula =√ 12/√3+√27/√3.
=√( 12/3)+√(27/3)
=√4+√9
=5.
Example 4, Simplification: √3/√8
Solution: √ 3/√ 8 = √ 3/2 √ 2 = (√ 3 * √ 2)/(2 * √ 2) = √ 6/4.
Another solution: √ 3/√ 8 = √ (3/8) = √ (3 * 2)/(8 * 2) = √ 6/16 = √ 6/4.
Example 3 is to remove the root sign by simplification, and example 4 is to simplify the real number formula with root sign by using the basic properties of fraction.
Example 5, Simplification: 1/(√3-√2)
Solution: Original formula =(√3+√2)/[(√3-√2)(√3+√2)]
=(√3+√2)/(3-2)
=√3+√2.
This topic uses the square difference formula and the basic properties of fractions to remove the root sign on the denominator.
4. Comprehensive application
(1), use √a≥0 and a≥0 to solve the problem.
In Example 6, it is known that √(x+5)+√(y+3)=0, so x-y is found.
Solution: √√(x+5)≥0, √(y+3)≥0 and √(x+5)+√(y+3)=0.
∴x+5=0,y+3=0
∴x=5,y=3.
∴x-y=-5-(-3)=-2.
In Example 7, it is known that y=√(x-2)+√(2-x)+4.
Looking for xy
Solution: ∫x-2≥0, 2-x≥0 ∴x=2.
y=4
∴xy=8.
Note: Example 5 uses the nonnegativity of the arithmetic square root, and Example 7 uses the nonnegativity of its square root.
(2) Comprehensive (flexible) application
Example 8, Simplification: (√6+4√3+3√2)/[(√6+√3)(√3+√2)]
Solution: The original formula = [(√ 6+√ 3)+3 (√ 3+√ 2)/[(√ 6+√ 3) (√ 3+√ 2).
= 1/(√3+√2)+3/(√6+√3)
=√3-√2+√6-√3
=√6-√3.
Example 9, Simplification: (8+2 √15-√10-6)/(√ 5+√ 3-√ 2)
Solution: The original formula = [5+2 √15+3-2 (√ 5+√ 3)]/(√ 5+√ 3-√ 2)
=[(√5+√3)-√2(√5+√3)]/(√5+√3-√2)
=[(√5+√3)(√5+√3-√2)]/(√5+√3-√2)
=√3+√3.
Examples 8 and 9 are complex questions with roots by comprehensively applying fractional properties and flexibly applying multiplication formula and distribution law (reverse application).