Two calculation problems of real number operation 3^1/2 - 27^1/2 ÷ 8^1/3 8 ^ 2 / 3 × 4 ^ 1 / 4 - radical (1-radical 2) ^ 2

Two calculation problems of real number operation 3^1/2 - 27^1/2 ÷ 8^1/3 8 ^ 2 / 3 × 4 ^ 1 / 4 - radical (1-radical 2) ^ 2

=3^1/2 -3^(3/2)÷(2^3)^(1/3)
=3^1/2 -3×3^(1/2)÷2
=-1/2 ×3^(1/2)
That is - √ 3 / 2
2、=(2^3)^(2/3)×(2^2)^(1/4) -|1-√2|
=2^2×2^(1/2)-√2+1
=4√2-√2+1
=3√2 +1

Operation of real number
evaluation
Root 6-root 35 (that is to find the arithmetic square root of 6-root 35)
Subtract root 6 + root 35 (that is, find the arithmetic square root of 6 + root 35)
Add the values of two items

Root 6 - root 35 = √ [(12-2 √ 35) / 2] = √ [(√ 7 - √ 5) ^ 2 / 2] = (√ 7 - √ 5) * √ (1 / 2) = (√ 14 - √ 10) / 2 similarly: root 6 + root 35 = (√ 14 + √ 10) / 2 root 6 - root 35 minus root 6 + root 35 = (√ 14 - √ 10) / 2 - (√ 14 + √ 10) / 2 = - √ 10 root 6 - root 35 + root 6 + root 35

Real number operation
√2(√2-√3)÷6=
1-³√ -8+√ 2*（-2/√ 8)=
(- 1) to the power of 2020 * 4 + 1 + | √ 3-2sin60|

1、=(2-√6)÷6
=1/3 -√6 /6
2、=1+2-1
=2
3、=1×4+1+|√3-√3|
=5

It is necessary to change the denominator in the equation x 0.3 = 1 + 1.2-0.3x 0.2 into an integer

x\0.3=1+1.2-0.3X\0.2
Turn 0.2 and 0.3 in the equation into denominators
Then multiply both sides of the equation by 10
So you have to
x/3=10+(1.2-0.3x)/2

What is the denominator in the equation 0.3 x = 1 + 0.2 x = 1.2-0.3x

x/0.3=1+1.2/0.2-0.3x
10x/3=1+6-3x/10

What's the correct way to remove the denominator from the equation 2 / X-6 / X-1 = 1

2 X-6 X-1 = 1 de denominator
3x-x-6=6
2x=12
x=6

If we solve the equation x + 3 / 3-x-1 / 6 = 5-x / 3 and remove the denominator, the result is correct

x+3/3-x-1/6=5-x/3,
2x+6-x+1=10-2x;
3x=3;
x=1;
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Given that the solution of equation 3 [x + a] - [x + 5] = 13 is twice the solution of equation 2 [x-4] + 3 = 6, find a

[X-4]+3=6
x-4=3
x=7
Therefore, the solution of 3 [x + a] - [x + 5] = 13 is 7 * 2 = 14
3[14+a]-[14+5]=13
42+3a-19=13
3a=-10
a=-10/3

To solve the equation: (x + 1) / (x + 2) + (x + 6) / (x + 7) = (x + 2) / (x + 3) + (x + 5) / (x + 6), why must the difference of denominator constant on both sides of the equation be equal?
My solution: the original equation=
（x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)
1-1/(x+2)+1-1/(x+7)=1-1/(x+3)+1-1/(x+6)
-1/(x+2)-1/(x+7)=-1/(x+3)-1/(x+6)
1/(x+2)+1/(x+7)=1/(x+3)+1/(x+6)
(2x+9)/(x^2+9x+14)=(2x+9)/(x^2+9x+18)
Why can't we find out in this way? We have to press
1/(x+2)-1/(x+3)=1/(x+6)-1/(x+7)
(x+3-(x+2))/(x+2)(x+3)=(x+7-(x+6))/(x+6)(x+7)
1/(x+2)(x+3)=1/(x+6)(x+7)
(x+2)(x+3)=(x+6)(x+7)
x^2+5x+6=x^2+13x+42
8x=-36
x=-9/2
That's right. It should work
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It can also be solved. (2x + 9) / (x ^ 2 + 9x + 14) = (2x + 9) / (x ^ 2 + 9x + 18) to extract the common factor: (2x + 9) [1 / (x ^ 2 + 9x + 14) - 1 / (x ^ 2 + 9x + 18)] = 0, then there is 2x + 9 = 0 = > x = - 9 / 2 or: 1 / (x ^ 2 + 9x + 14) - 1 / (x ^ 2 + 9x + 18) = 0, after dividing, there is 4 / [(x ^ 2 + 9x + 14) * (x ^ 2 + 9x + 18)] = 0

Equation 5-2: 1 + 2x = 0

Equation 5-2: 1 + 2x = 0
5-2 / 1 + 2x = 0
10-1+4x=0
9+4x=0
4x= -9
x= -9\4