26 uždavinys
Raskite didžiausią sveikąjį lygties $$\sqrt {x^{2}-4\cdot x+12} = 3$$ sprendinį.
Sprendimas.
saknis(
x^2- 4* x+12) =
3
x^2- 4* x+12) = 3
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saknis( x^2- 4* x+12) = 3$$\sqrt {x^{2}-4\cdot x+12}$$ = $$3$$
x^2- 4* x+12) = 3$$\sqrt {x^{2}-4\cdot x+12}$$ = $$3$$ (saknis( x^2- 4* x+12))^2 = 3^2$$(\sqrt {x^{2}-4\cdot x+12})^{2}$$ = $$3^{2}$$
x^2- 4* x+12))^2 = 3^2$$(\sqrt {x^{2}-4\cdot x+12})^{2}$$ = $$3^{2}$$$${\normalsize (\sqrt {x^{2}-4\cdot x+12})^{2}}$$ = $${\normalsize (x^{2}-4\cdot x+12)}$$
x^2- 4* x+12 = 3^2$$x^{2}-4\cdot x+12$$ = $$3^{2}$$
$${\normalsize 3^{2}}$$ = $${\normalsize 9}$$
x^2- 4* x+12 = 9$$x^{2}-4\cdot x+12$$ = $$9$$
x^2- 4* x+12-9 = 0$$x^{2}-4\cdot x+12-9$$ = $$0$$
$${\normalsize 12-9}$$ = $${\normalsize 3}$$
x^2- 4* x+3 = 0$$x^{2}-4\cdot x+3$$ = $$0$$
$${\normalsize x^{2}-4\cdot x+3}$$ = $${\normalsize (x-3)\cdot (x-1)}$$ Paaiškinimas:
Paaiškinimas: Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = -4, c = 3.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 16-12}$$ = 4.
x1 = $${\normalsize \frac{4+\sqrt {4}}{2\cdot 1} = \frac{4+2}{2} = \frac{2+1}{1} = \frac{3}{1}}$$ = 3
x2 = $${\normalsize \frac{4-\sqrt {4}}{2\cdot 1} = \frac{4-2}{2} = \frac{2-1}{1} = \frac{1}{1}}$$ = 1
a = 1, b = -4, c = 3.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 16-12}$$ = 4.
x1 = $${\normalsize \frac{4+\sqrt {4}}{2\cdot 1} = \frac{4+2}{2} = \frac{2+1}{1} = \frac{3}{1}}$$ = 3
x2 = $${\normalsize \frac{4-\sqrt {4}}{2\cdot 1} = \frac{4-2}{2} = \frac{2-1}{1} = \frac{1}{1}}$$ = 1
(x-3)* (x-1) = 0$$(x-3)\cdot (x-1)$$ = $$0$$
Gavome dvi šaknis: x = 1 ir x = 3.
Didesnė šaknis x = 3.
Patikriname lygybę, kai x = 3:
saknis( x^2- 4* x+12) =
3
x^2- 4* x+12) = 3
|
|
saknis( x^2- 4* x+12) = 3$$\sqrt {x^{2}-4\cdot x+12}$$ = $$3$$
x^2- 4* x+12) = 3$$\sqrt {x^{2}-4\cdot x+12}$$ = $$3$$ Paaiškinimas: Keitimas $${\normalsize x}$$ = $${\normalsize 3}$$.
saknis( 3^2- 4* 3+12) = 3$$\sqrt {3^{2}-4\cdot 3+12}$$ = $$3$$
3^2- 4* 3+12) = 3$$\sqrt {3^{2}-4\cdot 3+12}$$ = $$3$$$${\normalsize 3^{2}}$$ = $${\normalsize 9}$$
saknis(9- 4* 3+12) = 3$$\sqrt {9-4\cdot 3+12}$$ = $$3$$
9- 4* 3+12) = 3$$\sqrt {9-4\cdot 3+12}$$ = $$3$$$${\normalsize 4\cdot 3}$$ = $${\normalsize 12}$$
saknis(9-12+12) = 3$$\sqrt {9-12+12}$$ = $$3$$
9-12+12) = 3$$\sqrt {9-12+12}$$ = $$3$$$${\normalsize -12+12}$$ = $$0$$
saknis(9+0) = 3$$\sqrt {9+0}$$ = $$3$$
9+0) = 3$$\sqrt {9+0}$$ = $$3$$saknis(9) = 3$$\sqrt {9}$$ = $$3$$
9) = 3$$\sqrt {9}$$ = $$3$$$${\normalsize \sqrt {9}}$$ = $${\normalsize 3}$$
3 = 3$$3$$ = $$3$$
$$\sqrt {3^{2}-4\cdot 3+12}$$ = $$3$$
$$\sqrt {9-12+12}$$ = $$3$$
$$\sqrt {9}$$ = $$3$$
$$3$$ = $$3$$
Lygybė teisinga, taigi, didžiausia sveika lygties šaknis ir bus x = 3.
Atsakymas: 3