15 uždavinys
Išspręskite lygtį $$\sqrt {2-x} = x$$
Sprendimas.
saknis(
2-x) =
x
2-x) = x
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saknis(2-x) = x$$\sqrt {2-x}$$ = $$x$$
2-x) = x$$\sqrt {2-x}$$ = $$x$$ (saknis(2-x))^2 = x^2$$(\sqrt {2-x})^{2}$$ = $$x^{2}$$
2-x))^2 = x^2$$(\sqrt {2-x})^{2}$$ = $$x^{2}$$$${\normalsize (\sqrt {2-x})^{2}}$$ = $${\normalsize (2-x)}$$
2-x = x^2$$2-x$$ = $$x^{2}$$
2 = x^2+x$$2$$ = $$x^{2}+x$$
0 = x^2+x-2$$0$$ = $$x^{2}+x-2$$
$${\normalsize x^{2}+x-2}$$ = $${\normalsize (x-1)\cdot (x+2)}$$ Paaiškinimas:
Paaiškinimas: Kvadratinis trinaris $${\normalsize a\cdot x^{2}+b\cdot x+c}$$, kur
a = 1, b = 1, c = -2.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-8)}$$ = 9.
x1 = $${\normalsize \frac{-1+\sqrt {9}}{2\cdot 1} = \frac{-1+3}{2} = \frac{2}{2}}$$ = 1
x2 = $${\normalsize \frac{-1-\sqrt {9}}{2\cdot 1} = \frac{-1-3}{2} = \frac{-4}{2}}$$ = -2
a = 1, b = 1, c = -2.
Diskriminantas $${\normalsize D = b^{2}-4\cdot a\cdot c = 1-(-8)}$$ = 9.
x1 = $${\normalsize \frac{-1+\sqrt {9}}{2\cdot 1} = \frac{-1+3}{2} = \frac{2}{2}}$$ = 1
x2 = $${\normalsize \frac{-1-\sqrt {9}}{2\cdot 1} = \frac{-1-3}{2} = \frac{-4}{2}}$$ = -2
0 = (x-1)* (x+2)$$0$$ = $$(x-1)\cdot (x+2)$$
x-1 = 0$$x-1$$ = $$0$$
x = 1$$x$$ = $$1$$
x+2 = 0$$x+2$$ = $$0$$
x = -2$$x$$ = $$-2$$
Gavome x = 1 ir x = - 2, bet pastaroji šaknis netinka, nes $$\sqrt {2-(-2)}$$ ≠ $$-2$$
Atsakymas: x = 1