How to add, subtract, multiply and divide with root sign?

How to add, subtract, multiply and divide with root sign?

A radical cannot be added or subtracted directly, unless it is the same radical. Multiplication and division is very simple. Just multiply and divide the number under the root sign and add a root sign

All algorithms of radical

The number under the square root must be greater than or equal to 0; but if it is the third power root, it can be a negative number, so the specific situation of the specific analysis!
The following is the solution as square root
Add or subtract: there is no other method, only use calculator to calculate the specific value and then add or subtract;
When multiplying: the multiplication of two numbers with square root is equal to the product of two numbers under the root sign;
Division: the division of two numbers with square root will be equal to the quotient of two numbers under the root sign, and then simplify;
Then, sometimes if the denominator is a radical formula, we will choose to rationalize it so that the denominator has no root and the root is transferred to the molecule

How to add, subtract, multiply and divide Radix? Because I haven't learned it, if you want to have more insight, how to add, subtract, multiply and divide the numbers outside or inside the root sign? It's best to give an example to illustrate the white point Why is 8 equal to 2

You can't add or subtract directly, unless the numbers under the root sign are the same, for example, 2 (root 3) - (root 3) = root 3, which can be multiplied and divided by (root 2) × (root 3) = root 6

We know that the circumference of RT △ ABC is 4 + 4 root sign 3, and the center line of hypotenuse is 2. Find the area of △ ABC

The three sides of RT △ ABC are a, B, C
a+b+c=4+4√3
Bevel C = 2 * 2 = 4
a+b=4√3 ①
a²+b²=c²=16 ②
①²－②
2ab=32
ab=16
Δ ABC area = AB / 2 = 16 / 2 = 8

If the median length of the hypotenuse in RT △ ABC is 3 / 2 and the circumference of the triangle is 3 + 2, then the area of the triangle is?

Because the center line of the hypotenuse in RT △ ABC is equal to half of the hypotenuse, the hypotenuse AC = 2 * 3 / 2 = 3
So AB + BC = 3 + 2 √ 3-3 = 2 √ 3
So (AB + BC) ^ 2 = (2 √ 3) ^ 2 = 12
According to Pythagorean theorem, AB ^ 2 + BC ^ 2 = AC ^ 2 = 9
So the area of RT △ ABC = 1 / 2Ab * BC = 1 / 2 * 1 / 2 [(AB + BC) ^ 2 - (AB ^ 2 + BC ^ 2)] = 1 / 4 (12-9) = 3 / 4

Given that the circumference of RT triangle ABC is 4 + 4, root sign 3, and the center line on the hypotenuse is 2, then s triangle ABC =?

Let the two right angles of a triangle be a and B, and the hypotenuse of a triangle be c
A + B + C = 4 + 4 root sign 3;
Since the length of the center line on the bevel is 2, C = 2 * 2 = 4;
So a + B = 4 root sign 3;
Because it is a right triangle, so a square + b square = 16, that is, (a + b) square - 2Ab = 16, (4 root sign 3) square - 2Ab = 16;
The triangle area = 1 / 2Ab=

In the right angle △ ABC, the length of the hypotenuse is 2 and the circumference is 2+ 6, then the area of △ ABC is______ .

Let the hypotenuse of right angle △ ABC be C, and the two right angle sides be a and B,
According to the meaning of the title, a + B is obtained=
6，a2+b2=c2=4，
Then the area of △ ABC = 1
2ab=1
4[（a+b）2-（a2+b2）]=1
4（6-4）=1
2．
So the answer is 1
2．

The hypotenuse length of the right triangle ABC is 2, the perimeter is 3 + the root 3 Please get it out by four o'clock this afternoon,

Let two right angle sides be A.B, that is, s = 1 / 2 * sin90 * ab. let a + B = 4, a + B + 2 = 3 + radical 3
(a + b) square = (3 + radical 3) square expansion has a square + 2Ab + B = 4 + 2 * root sign 3. If a + B = 4 is brought into the above formula, 2Ab = 2 * root 3 is obtained, so AB = root 3, s = 1 / 2 * root 3

Given that the circumference of the right triangle ABC is 4 + 2, the root sign 2, find the maximum area of the triangle

Let the three sides of a triangle be a, B, C, where C is a hypotenuse and the perimeter is l, so there is L = a + B + C = a + B + √ (a ^ 2 + B ^ 2) because a + B ≥ 2 √ (AB), √ (a ^ 2 + B ^ 2) ≥ √ (2Ab), so l ≥ 2 √ (AB) + √ (2Ab) can be obtained by substituting s = AB / 2 into s ≤ L ^ 2 / [4 (3 + 2 √ 2)] that is, a right triangle with a fixed perimeter L

If the circumference of a right triangle ABC is 6 + 2 * (root 3), and the length of the central line on the hypotenuse is 2, then the area of the triangle =? The answer is 2 * (radical 3)

The center line on the hypotenuse is half of the hypotenuse
Therefore, the length of beveled edge is 4
Then the sum of the lengths of the two right angles is 6 + 2 √ 3-4 = 2 + 2 √ 3
If one right angle side length x, the other right angle side length is 2 + 2 √ 3-x
From the title: x ^ 2 + (2 + 2 √ 3-x) ^ 2 = 4 ^ 2
The solution is: X1 = 2, X2 = 2 √ 3
After calculation, the two right angles are 2 and 2 √ 3 respectively
So the area is 2 * 2 √ 3 / 2 = 2 √ 3