SOLVE EQUATION X 2 X SQUARED 1 2X SQUARED 1 X 2

Hello: Let x (2x -1)=y

y+1/y=2

y²+1=2y

y²-2y+1=0

y-1)²=0

y=1x/(2x²-1)=0

2x²-1=x

2x²-x-1=0

2x+1)(x-1)=0

x1=-1/2 x2=1

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Solution: Let x (2x -1)=y

Then the original formula can be reduced to y+1 y=2

Sorted out y -2y + 1 = 0, solved y = 1 (ps: I won't write the test) x (2x -1) = 1

2x -x-1=0

The solution is x = 1, x =

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Solution: x (2x -1)+(2x -1) x=2, let :2x -1=y, then the original formula is:

x-y) =0, then: x=y. Solution:

x-(2x -1)] =0, which is reduced to: 2x -x-1=0, and the solution is: x1=1, x2=-1 2.

Summary. Hello! Dear, is your question (x-5) =2x-1) squared? <>x-5) = 2x-1) + solve the equation with direct squares.

Hello! Dear, is your question (x-5) =2x-1) squared? <>x-5)²=2x-1)²

Hello! Dear, the solution process has been sent to you, directly open the flat method, we want to make the quadratic equation into the form of x = a, and then open the square on both sides of the equal sign at the same time, this problem we take the equal sign after the square (2x -1) as a whole, and then open the square on both sides, the right side of the equal sign is, (2x-1), will you? Pro-<>

x-1）²=2

x-1=±√2

x=±√2+1

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x-1)=2(x-1) (x-1)-2(x-1)=0 (x-1)(x-3)=0 x-1=0 or x-3=0 x=1 or x=3 If you are satisfied, please smile in time.

Oh. x-1) squared - 2 (x-1) - 3 = 0

x-1-3)(x-1+1)=0

x-4)x=0

x1=0x2=4

x-1)^2-2(x^2-1)=0

x^2-2x+1-2x^2+2=0

x^2+2x-3=0

x+3)(x-1)=0

x=1 or x=-3

x 1) 2 squared contains 4

x 1) 2 squared -4 = 0 (x-1 + 2) (x-1-2) = 0 (x + 1) (x-3) = 0x1 = -1 x2 = 3

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x-1) squared -2 (x-1) = 15

x-1)²-2(x-1)-15=0

x-1-5)(x-1+3)=0

x-6)(x+2)=0

x1=6 x2=-2

x²/（x-1）-2=x

x²-2（x-1）=x²-x

x²-2x+2=x²-x

The square of x = 2x minus two x minus x minus 1 the absolute value minus 1 is equal to zero.

x²-2x-|x-1|-1=0

x²-2x+1-|x-1|-2=0

x-1)²-x-1|-2=0

x-1|²-x-1|-2=0

Order|x-1|=t>=0

t²-t-2=0

t-2)(t+1)=0

t=2, t=-1 rounded.

x-1|=2

x-1=2 or x-1=-2

x=3 or x=-1

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x-1)²=2x-1)²

2x-1)²-x-1)²=0

2x-1)-(x-1)][2x-1)+(x-1)]=0

x(3x-2)=0

x1 = 0, x2 = 2/3

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Your original question is x -2x-|x-1|-5=0, right?

It can be written as (x-1) -x-1|-6=0

x-1|²-x-1|-6=0

x-1|-3)(|x-1|+2)=0

Will|x-1|Look at it as a whole.

The above formula gets, |x-1|=3 or |x-1|=-2 (rounded) so |x-1|=3

So x-1=3 or x-1=-3

So x1 = 4 and x2 = -2

(1) 3 (2y + 1) square = 27

2y+1)^2=9

2y+1=3 or 2y+1=-3

y=1 or y=-2

2) 4 (x-2) square = (x + 1) square.

At the same time, the prescription is prescribed. 2(x-2)=x+1 or 2(x-2)=-(x+1)x=5 or x=1

Hello: (1+x) squared + 2(1+x)-4=0(1+x) squared +2(1+x)+1=5

x+1+1)²=5

x+2=±√5

x=-2±√5

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Solution: (1+x) squared + 2(1+x)-4=0(1+x) squared + 2(1+x)+1=5

1+x+1)²=5

x+2)²=5

x+2 = 5 or x+2=- 5

i.e.: x=-2+5 or x=-2-5

I wish you progress in your studies and a happy life!

2x+1) = (x-1) squared.

2x+1)²-x-1)²=0

2x+1+x-1)(2x+1-x+1)=03x(x+2)=0

x=0 or x=-2

x squared -2x+1=49

x²-2x-48=0

x-8)(x+6)=0

x=8 or x=-6

2x+1) = (x-1) squared.

2x+1)^2-(x-1)^2=0

2x+1+x-1)(2x+1-x+1)=03x*(x+2)=0

3x=0 x+2=0

x=0 x=-2

x sells or flattens the wheel square - 2x + 1 = 49

x-1)^2=49

x-1=7 x-1=-7

x=8 x=-6

What to solve? Solving equations?

Solution 1: (2x+1) squared = (x-1) squared (2x+1) 2=(x-1) 2

4x^2+4x+1=x^2-2x+1

3x^2+6x=0

3x(x+2)=0

Solution: x1=0, x2=-2

Or: (2x+1) 2=(x-1) 2

2x+1=±(x-1)

Yes: 2x+1=x-1.........1)

Or: 2x+1=1-x.........2)

From (1): x=-2

From (2): 3x=0, i.e.: x=0

That is, the solution of the original equation is: x1=-2, x2=0

Solution 2: The square of x - 2x + 1 = 49

x^2-2x+1=49

x+1)^2=7^2

x+1=±7

x=1±7x1=8、x2=-6