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Raskite didžiausią sveikąjį lygties x24x+12=3\sqrt {x^{2}-4\cdot x+12} = 3 sprendinį.

Sprendimas.

saknis( x^2- 4* x+12)  = 
3
saknis( x^2- 4* x+12) = 3x24x+12\sqrt {x^{2}-4\cdot x+12} = 33
 (saknis( x^2- 4* x+12))^2 =  3^2(x24x+12)2(\sqrt {x^{2}-4\cdot x+12})^{2} = 323^{2}
(x24x+12)2{\normalsize (\sqrt {x^{2}-4\cdot x+12})^{2}} = (x24x+12){\normalsize (x^{2}-4\cdot x+12)}
 x^2- 4* x+12 =  3^2x24x+12x^{2}-4\cdot x+12 = 323^{2}
32{\normalsize 3^{2}} = 9{\normalsize 9}
 x^2- 4* x+12 = 9x24x+12x^{2}-4\cdot x+12 = 99
 x^2- 4* x+12-9 = 0x24x+129x^{2}-4\cdot x+12-9 = 00
129{\normalsize 12-9} = 3{\normalsize 3}
 x^2- 4* x+3 = 0x24x+3x^{2}-4\cdot x+3 = 00
x24x+3{\normalsize x^{2}-4\cdot x+3} = (x3)(x1){\normalsize (x-3)\cdot (x-1)}
Paaiškinimas:
Kvadratinis trinaris ax2+bx+c{\normalsize a\cdot x^{2}+b\cdot x+c}, kur
a = 1, b = -4, c = 3.
Diskriminantas D=b24ac=1612{\normalsize D = b^{2}-4\cdot a\cdot c = 16-12} = 4.
User posted image
x1 = 4+421=4+22=2+11=31{\normalsize \frac{4+\sqrt {4}}{2\cdot 1} = \frac{4+2}{2} = \frac{2+1}{1} = \frac{3}{1}} = 3
x2 = 4421=422=211=11{\normalsize \frac{4-\sqrt {4}}{2\cdot 1} = \frac{4-2}{2} = \frac{2-1}{1} = \frac{1}{1}} = 1
 (x-3)* (x-1) = 0(x3)(x1)(x-3)\cdot (x-1) = 00


Gavome dvi šaknis: x = 1 ir x = 3.

Didesnė šaknis x = 3.

Patikriname lygybę, kai x = 3:

saknis( x^2- 4* x+12)  = 
3
saknis( x^2- 4* x+12) = 3x24x+12\sqrt {x^{2}-4\cdot x+12} = 33
Paaiškinimas:
Keitimas x{\normalsize x} = 3{\normalsize 3}.
saknis( 3^2- 4* 3+12) = 33243+12\sqrt {3^{2}-4\cdot 3+12} = 33
32{\normalsize 3^{2}} = 9{\normalsize 9}
saknis(9- 4* 3+12) = 3943+12\sqrt {9-4\cdot 3+12} = 33
43{\normalsize 4\cdot 3} = 12{\normalsize 12}
saknis(9-12+12) = 3912+12\sqrt {9-12+12} = 33
12+12{\normalsize -12+12} = 00
saknis(9+0) = 39+0\sqrt {9+0} = 33
saknis(9) = 39\sqrt {9} = 33
9{\normalsize \sqrt {9}} = 3{\normalsize 3}
3 = 333 = 33
3243+12\sqrt {3^{2}-4\cdot 3+12}  = 33
912+12\sqrt {9-12+12}  = 33
9\sqrt {9}  = 33
33  = 33

Lygybė teisinga, taigi, didžiausia sveika lygties šaknis ir bus x = 3.

Atsakymas: 3

25 uždavinys27 uždavinys