This document contains solutions to three equations involving fractions equal to values. The first equation, x^2 + x - 2 = 0, is solved to find the values of x that satisfy the equation, which are x = 1 and x = -2. The second equation, (x-1)/(x+1) = 3, is solved to find the value of x that makes the fraction equal to 3, which is x = -2. The third equation, x/(x-1) - 1/(x+2) - 3/(x-1)(x+2) = 0, is solved but does not have any real number solutions for x.

1. Sudėtingesnės
trupmeninės lygtys

3. Pvz.:
x2 + x − 2
=0
x −1
x2 + x − 2 = 0
D = 12 − 4 ⋅1 ⋅ (−2) = 1 + 8 = 9
−1 + 9 −1 + 3 2
−1− 9 −1− 3 − 4
x1 =
=
= = 1; x2 =
=
=
= −2;
2 ⋅1
2
2
2 ⋅1
2
2
Kai x = 1, tai x – 1 = 1 – 1 = 0 (atmetame);
Kai x = -2, tai x – 1 = -2 – 1 = -3.
Ats.: x = -2.

4. x −1
Raskime x reikšmes su kuriomis trupmenos
x +1
reikšmė lygi 3.
x −1
=3
Sprendžiame lygtį:
x +1
x −1
−3 = 0
⇒
x +1
x − 1 3( x + 1)
−
=0
x +1
x +1
x − 1 − 3x − 3
=0
x +1
x −1 3
− =0
x +1 1
x − 1 − 3( x + 1)
⇒
=0
x +1
− 2x − 4
⇒
=0
x +1

5. Gavome lygtį:
− 2x − 4
=0
x +1
Trupmena lygi 0, kai:
− 2 x − 4 = 0 ⇒ − 2( x + 2) = 0 ⇒ x + 2 = 0, x = −2;
Kai x = -2, tai x +1 = -2 + 1 = -1 (tinka);
Ats.: x = -2.

6. Išspręskime lygtį:
x
1
3
−
−
=0
x − 1 x + 2 ( x − 1)( x + 2)
x( x + 2) − ( x − 1) − 3
=0
( x − 1)( x + 2)
x + 2x − x +1− 3
=0
( x − 1)( x + 2)
2
x + x−2
=0
( x − 1)( x + 2)
2

7. x + x−2
=0
( x − 1)( x + 2)
2
Gavome lygtį:
Trupmena lygi 0, kai:
x2 + x − 2 = 0
D = b 2 − 4ac = 12 − 4 ⋅1 ⋅ (−2) = 1 + 8 = 9
− b − D −1− 9 −1− 3 − 4
x1 =
=
=
=
= −2
2a
2 ⋅1
2
2
− b + D −1 + 9 −1 + 3 2
x2 =
=
=
= =1
2a
2 ⋅1
2
2

8. Kai x = -2, tai ( x – 1)( x +2) =
= (-2– 1)( -2 +2) = -3·0 = 0 (netinka);
Kai x = 1, tai ( x – 1)( x +2) =
= (1 – 1)(1 +2) = 0·3 = 0 (netinka);
Ats.: Sprendinių nėra.