Substitution can be used in the following topics, that is, 2∧x and the like can be regarded as a.

2∧(2x)-2∧x *(ì2+ 1)+√2 = 0

Solution: (2 ∧ x) ∧ 2-2 ∧ x * (∧ 2+1)+√ 2 = 0.

(2∧x- 1)(2∧x-√ 2) = 0

Therefore: 2 ∧ x- 1 = 0 2 ∧ x-√ 2 = 0.

X=0 x= 1/2

2∧(2x)+2∧x *(4–√2)-4 *√2 = 0

Solution: (2 ∧ x) ∧ 2+2 ∧ x * (4 –√ 2)-4 * √ 2 = 0.

(2∧x+4)( 2∧x-√ 2) = 0

2∧x+4=0 2∧x-√ 2=0

2∧x+4=0 (no solution in real number range) x= 1/2.

3∧(2x)-3∧x *(ì3+9)+9 *√3 = 0

Solution: (3 ∧ x) ∧ 2-3 ∧ x * (∧ 3+9)+9 * √ 3 = 0.

(3∧x-9)(3∧x-√3)=0

3∧x-9=0 3∧x-√3=0

X=2 x= 1/2

3∧(2x)- 1-3∧x *(ì3+ 1)+3 *√3 = 0

Solution: (3 ∧ x) ∧ 2-3 ∧ x * (√ 3+1)+3 * √ 3-1= 0.

There seems to be something wrong with the title.

2∧x+ 10+2∧(4-x)=0

Solution: 2∧x+ 10+2∧4/(2∧x)=0.

(2∧x)∧2+ 10×2∧x+2∧4=0

(2∧x+2)(2∧x+8)=0

No solution in real number range

[The estimated topic is 2 ∧ x- 10+2 ∧ (4-x) = 0, then (2 ∧ x-2) (2 ∧ x-8) = 0, so: x =1x =

2∧x-36+2∧(7-x)=0

Solution: 2∧x-36+2∧7/(2∧x)=0.

(2∧x)∧2-36×2∧x+2∧7=0

(2∧x-2∧5)( 2∧x-2∧2)=0

2∧x-2∧5=0 2∧x-2∧2=0

X=5 x=2

(3/2)∧x-(2/3)∧( x+ 1)= 2 1 1/ 108

Solution: (3/2) ∧ x-2/3×/[(3/2) ∧ x]-21108 = 0.

[(3/2)∧x]∧2-2 1 1/ 108×[(3/2)∧x)-2/3 = 0

[(3/2)∧x+486/ 108][(3/2)∧x-64/ 108]= 0

(3/2)∧x+486/ 108=0 (no solution in real number range) (3/2)∧x-64/ 108=0.

X=-4

(5/3)∧(x- 1)/(3/5)∧(x-2)= 125/27

Solution: (5/3) ∧ (x-1)/(3/5) ∧ (x-2) =125/27.

[(5/3)∧(x- 1)][(5/3)∧(x-2)]= 125/27

(x- 1)+(x-2)=3

X=3