Задачи с корени и уравнение

[Георги Иванов]

Моля за помощ! може да ми ги пратите на емайл - gotkit@abv.bg

[mail_dinko]

Задача 1
[tex]\sqrt {2}.\sqrt {3}.\sqrt {10}.\sqrt {15}=[/tex]
[tex]= \sqrt {2.3.10.15} = \sqrt {2.3.2.5.3.5}=2.3.5= 30[/tex]

Задача 2
[tex]\frac {\sqrt {7}. \sqrt {63}}{\sqrt {5} . \sqrt {20}} = \frac {\sqrt {7.7.3^2}}{5.5.4}=\frac {21}{20}[/tex]

Задача 3
[tex](3 \sqrt {2} - \sqrt {3} )^2 - \sqrt {3} ( \sqrt {2} - 1)=[/tex]
[tex]= 18+ 3-6 \sqrt {6} - \sqrt {6} + \sqrt {3}= 21 - 7 \sqrt {6} + \sqrt {3}[/tex]

Задача 4
[tex]3x^2+5x=0[/tex]
[tex]x(3x+5)=0[/tex]
[tex]x_1=0[/tex]
[tex]x_2= - \frac53[/tex]

Задача 5
[tex]4x^2-7=0[/tex]
[tex]4x^2 =7|:4[/tex]
[tex]x_{1,2}= \pm \frac {\sqrt {7}}{2}[/tex]

Задача 6
[tex]3x^2 - 4x + 1 =0[/tex]
[tex]D= 4 - 3= 1[/tex]
[tex]x_ {1,2}= \frac {2 \pm 1}{3}[/tex]
[tex]x_1 = 1[/tex]
[tex]x_2 = \frac13[/tex]

Задача 7
[tex](2x-1)(x+3)-(x+1)^2=6[/tex]
[tex]2x^2+ 6x -x-3 -x^2-1 -2x - 6=0[/tex]
[tex]x^2+3x-10=0[/tex]
[tex]D= 9 +4.10 = 49=7^2[/tex]
[tex]x_{1,2} = \frac {- 3 \pm 7}{2}[/tex]
[tex]x_1 = -5[/tex]
[tex]x_2 = 2[/tex]

[Davids]

1. $\sqrt{2}.\sqrt{3}.\sqrt{10}.\sqrt{15} = \sqrt{2.3.10.15} = \sqrt{2.3.2.5.3.5} = \sqrt{2^2.3^2.5^2} = 2.3.5 = 30$
2. $\frac{\sqrt{7}.\sqrt{63}}{\sqrt{5}.\sqrt{20}} = \sqrt{\frac{7.7.9}{5.5.4}} = \frac{7.3}{5.2} = \frac{21}{10}$
3. $(3\sqrt{2}-3)^2 - \sqrt{3}(\sqrt{2}-1) = 9.2 - 2.3.\sqrt{2.3} + 3 - \sqrt{3}.\sqrt{2} + \sqrt{3} = 21 - 7\sqrt{6} + \sqrt{3}$
4. $3x^2 + 5x = x(3x+5) = 0$
$x_1 = 0; x_2 = -\frac{5}{3}$
5. $4x^2 = 7$
$x^2 = \frac{7}{4}$
$x = \pm \sqrt{\frac{7}{4}} = \pm \frac{\sqrt{7}}{2}$
6. $3x^2 - 4x + 1 = 0$
$D = 4^2 - 4.3 = 4 = 2^2$
$x_1 = \frac{4-2}{2.3} = \frac{1}{3}$
$x_2 = \frac{4+2}{2.3} = 1$
7. $(2x-1)(x+3) - (x+1)^2 = 6$
$2x^2 + 6x - x - 3 - (x^2 + 2x + 1) - 6 = 0$
$2x^2 + 5x - 9 - x^2 - 2x - 1 = 0$
$x^2 + 3x - 10 = 0$
$D = 3^2 - 4.1.(-10) = 9 + 40 = 7^2$
$x_1 = \frac{-3-7}{2} = -5$
$x_2 = \frac{-3+7}{2} = 2$