To solve the system of bivariate linear equations with root sign: √ 3x - √ 2Y = 1 √ 2x - √ 3Y = 0 Ambiguity! The numbers under the radical sign of √ 3 x - √ 2 y = 1 √ 2 x - √ 3 y = 0 and X, y are separated. It's not a combination

To solve the system of bivariate linear equations with root sign: √ 3x - √ 2Y = 1 √ 2x - √ 3Y = 0 Ambiguity! The numbers under the radical sign of √ 3 x - √ 2 y = 1 √ 2 x - √ 3 y = 0 and X, y are separated. It's not a combination

√(3x)-√(2y)=1 ...(1)
√(2x)-√(3y)=0 ...(2)
From (2), we can get: √ (2x) = √ (3Y), √ y = √ (2x / 3), and bring in (1) to get:
√(3x)-√(2*2x/3)=1
Multiply both sides by √ 3:
3√x-2√x=√3
√x=√3
X=3
√y=√(2x/3)=√(2*3/3)=√2
Y=2
[√ 3 x - √ 2, y = 1 √ 2 x - √ 3, y = 0, the numbers under the radical sign and X, y are separated. They are not combined]
√3 x-√2 y=1 ...（1）
√2 x-√3 y=0 ...（2）
(1) * √ 3 - (2) * √ 2 is obtained as follows:
3x-2x=√3
x=√3
y=√2x/√3=√2*√3/√3=√2

Solving equations 2x radical 3Y = 0 radical 3x-2y = 1

2X radical 3Y = 0
x=√3y/2
Radical 3x-2y = 1
√3*√3y/2-2y=1
3y/2-2y=1
-y/2=1
y=-2
x=-√3

Solving equations: root 3x minus root 2Y = 1 ① root 2x minus root 3Y = 0 ②

If √ 2x - √ 3Y = 0 → → √ 2x = √ 3Y = → → 2x = 3Y → → x = 3Y / 2, then √ (9y / 2) - √ 2Y = 1 is the same square on both sides of the equation, then 9y / 2 + 2Y - √ (9y ^ 2) = 1, that is, 9y / 2 + 2y-3y = 1, that is, 7Y / 2 = 1, and the solution is y = 2 / 7 ℅ x = 3Y / 2 = 3 / 7

To solve the equations: ① root 2x + radical 3Y = 1; ② radical 3x + Radix 2Y = 2

① (3) 2
√3(√2x+√3y) - √2(√3x-√2y)=√3-2√2
That is, y = √ 3 - 2 √ 2
Substituting it into (1), we get the following result:
x=2√3-√2

Root 2x - radical 3Y = radical 6 radical 3x + radical 2Y = 2 radical 6

√2x-√3y=√6 ,(1)√3x+√2y=2√6 (2)(1)*√3= ==> √6x-3y=√18 ==> √6x=√18 +3y(2)*√2= ==> √6x +2y =2√12 ====>√6x =2√12 -2y√18+3y=2√12-2y3y+2y=2*2√3-3√25y=4√3-3√2y=(4√3-3√2)/5√2x-√3y=√...

If M satisfies the relation root 3x + 5y-m + root 2x + 3y-m = radical x-199 + y * radical 199-x-y, explore the value of M

From the last two root signs x-199 + y, 199-x-y, if you want to have meaning, you can only have X + y = 199. Therefore, the time can be changed into root sign (3x + 5y-m) + root sign (2x + 3y-m) = 0, then it is the solution
3x+5y-m=0
x+y=199
2x+3y-m=0
M = 199

If the real number m satisfies: root 3x + 5y-2-m + radical 2x + 3y-m = radical x + y-199 + radical 199-x-y, find the value of M

Radical x + y-199 > = 0
Radical 199-x-y > = 0
So, x + y-199 = 0, x + y = 199
Radix 3x + 5y-2-m + Radix 2x + 3y-m = = 0
3X+5Y-2-m=0===>3(x+y)+2y-2-m=0===>2y-m=-595
2X+3Y-m=0>y-m=-398
The solution is: M = 201

It is known that the quadratic equation of one variable (m-radical 2) x square + 3x + m square has a root of 0. Find the value of M

X = 0 satisfies the equation (M - √ 2) x? + 3x + m? = 0
M 2 = 0
So m = 0

The square + root sign 3x + 3 / 4 of the quadratic equation x = 0 The square of factorization (5-x) - 16 = 0

(5-x)²-16=0
x²-10x+25-16=0
x²-10x+9=0
(x-9)(x-1)=0
X = 9 or x = 1

X square - 9 = 2 radical sign 3x to solve quadratic equation of one variable

Xsquare-9 = 2 root sign 3x
x^2-2x√3-9=0
x=(2√3+-4√3)/2
x1=3√3 x2=-√3