24 uždavinys
Su kuriomis kintamojo x reikšmėmis reiškinio $$\frac{2}{x+1}$$ skaitinė reikšmė keturis kartus mažesnė už reiškinio $$\frac{x+1}{2}$$ skaitinę reikšmę?
Sprendimas.
| 2 |
| / (x+1) |
| (x+1) |
| / 2 |
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| 2 |
| / (x+1) |
| (x+1) |
| / 2 |
$${\normalsize \frac{2}{x+1}\cdot 4}$$ = $${\normalsize \frac{8}{x+1}}$$
| 8 |
| / (x+1) |
| (x+1) |
| / 2 |
8 =
$$8$$ = $$\frac{(x+1)\cdot (x+1)}{2}$$
| (x+1)* (x+1) |
| / 2 |
8* 2 = (x+1)* (x+1)$$8\cdot 2$$ = $$(x+1)\cdot (x+1)$$
$${\normalsize 8\cdot 2}$$ = $${\normalsize 16}$$
16 = (x+1)* (x+1)$$16$$ = $$(x+1)\cdot (x+1)$$
$${\normalsize (x+1)\cdot (x+1)}$$ = $${\normalsize (x+1)^{2}}$$
16 = (x+1)^2$$16$$ = $$(x+1)^{2}$$
(x+1)^2 = 16$$(x+1)^{2}$$ = $$16$$
$$\frac{2}{x+1}\cdot 4$$ = $$\frac{x+1}{2}$$
$$\frac{8}{x+1}$$ = $$\frac{x+1}{2}$$
$$8\cdot 2$$ = $$(x+1)\cdot (x+1)$$
$$16$$ = $$(x+1)^{2}$$
$$(x+1)^{2}$$ = $$16$$
Iš abiejų pusių traukiame šaknį, gauname dvi lygtis
$$\sqrt {(x+1)^{2}} = \sqrt {16}$$
ir
$$\sqrt {(x+1)^{2}} = -\sqrt {16}$$
saknis( (x+1)^2) =
saknis(16)
(x+1)^2) = saknis(
16)|
|
saknis( (x+1)^2) = saknis(16)$$\sqrt {(x+1)^{2}}$$ = $$\sqrt {16}$$
(x+1)^2) = saknis(16)$$\sqrt {(x+1)^{2}}$$ = $$\sqrt {16}$$$${\normalsize \sqrt {(x+1)^{2}}}$$ = $${\normalsize (x+1)}$$
x+1 = saknis(16)$$x+1$$ = $$\sqrt {16}$$
16)$$x+1$$ = $$\sqrt {16}$$$${\normalsize \sqrt {16}}$$ = $${\normalsize 4}$$
x+1 = 4$$x+1$$ = $$4$$
x = 4-1$$x$$ = $$4-1$$
$${\normalsize 4-1}$$ = $${\normalsize 3}$$
x = 3$$x$$ = $$3$$
$$\sqrt {(x+1)^{2}}$$ = $$\sqrt {16}$$
$$x+1$$ = $$4$$
$$x$$ = $$3$$
saknis( (x+1)^2) =
-saknis(16)
(x+1)^2) = -saknis(
16)|
|
saknis( (x+1)^2) = -saknis(16)$$\sqrt {(x+1)^{2}}$$ = $$-\sqrt {16}$$
(x+1)^2) = -saknis(16)$$\sqrt {(x+1)^{2}}$$ = $$-\sqrt {16}$$$${\normalsize \sqrt {(x+1)^{2}}}$$ = $${\normalsize (x+1)}$$
x+1 = -saknis(16)$$x+1$$ = $$-\sqrt {16}$$
16)$$x+1$$ = $$-\sqrt {16}$$$${\normalsize \sqrt {16}}$$ = $${\normalsize 4}$$
x+1 = -4$$x+1$$ = $$-4$$
x = -4-1$$x$$ = $$-4-1$$
$${\normalsize -4-1}$$ = $${\normalsize -5}$$
x = -5$$x$$ = $$-5$$
$$\sqrt {(x+1)^{2}}$$ = $$-\sqrt {16}$$
$$x+1$$ = $$-4$$
$$x$$ = $$-5$$
Atsakymas: x = -5 ir x = 3