Все эти задания решаются по теореме Виета.

1) х₁ + х₂ = - 3 + 5 = 2
х₁х₂ = - 3 * 5 = - 15
х² - 2 - 15 = 0

3) х₁ + х₂ = (3а + 1) + (5а - 2) = 3а + 1 + 5а - 2 = 8а - 1
х₁х₂ = (3а + 1)(5а - 2) = 15а² + 5а - 6а - 2 = 15а² - а - 2
х² - (8а - 1)х + (15а² - а - 2) = 0

5) х₁ + х₂ = (√7 - √6) + (√7 + √6) = 2√7
х₁х₂ = (√7 - √6)(√7 + √6) = √7² - √6² = 7 - 6 = 1
х² - 2√7 + 1 = 0

0 0

3 года назад

Решите пожалуйста 1 пример!! Даю 20 баллов!

danilka211

$\sqrt{x+2\sqrt{x-1}} -\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1+2\sqrt{x-1}+1} -\\\\-\sqrt{x-1-2\sqrt{x-1}+1}= \sqrt{(\sqrt{x-1})^2+2\cdot\sqrt{x-1}\cdot 1+1^2} -\\ \\ - \sqrt{(\sqrt{x-1})^2-2\cdot\sqrt{x-1}\cdot 1+1^2}=\sqrt{(\sqrt{x-1}+1)^2} -\\\\ -\sqrt{(\sqrt{x-1}-1)^2} =|{\sqrt{x-1}+1}|-|{\sqrt{x-1}-1}|=\\\\=\sqrt{x-1}+1-|{\sqrt{x-1}-1}|=\left \{ {{\sqrt{x-1}+1-(\sqrt{x-1}-1), \sqrt{x-1}-1\geq0} \atop {{\sqrt{x-1}+1-(1-\sqrt{x-1}), \sqrt{x-1}-1<0}} \right. =$

$=\left \{ {{\sqrt{x-1}+1-\sqrt{x-1}+1, \sqrt{x-1}\geq1} \atop {{\sqrt{x-1}+1-1+\sqrt{x-1}, \sqrt{x-1}<1}} \right.=\left \{ {{2, x-1\geq1} \atop {{2\sqrt{x-1}, 0\leq x-1<1}} \right. =\left \{ {{2, x\geq2} \atop {{2\sqrt{x-1}, 1\leq x<2}} \right.$

0 0

4 года назад

X^2-8x+15=0 X^2+12x+20=0 X^2-5x-6=0 X^2-8x-9=0 Найдите корни уравнения

Iuliia [1]

$x^2 - 8x + 15 = 0

 \left \{ {{x1+x2 = -p} \atop {x1*x2 = q}} \right. 

 \left \{ {{x1+x2 = 8} \atop {x1*x2 = 15}} \right. 

 \left \{ {{x1 = 5} \atop {x2 = 3}} \right.$

$x^2 + 12x + 20 = 0

D = 64

x1 = - 2 ; x2 = - 10$

$x^2 - 5x - 6 = 0

 \left \{ {{x1+x2 = -p} \atop {x1*x2 = q}} \right. 

 \left \{ {{x1+x2 = 5} \atop {x1*x2 = -6}} \right. 

 \left \{ {{x1 = 6} \atop {x2 = -1}} \right.$

$x^2 - 8x - 9 = 0

D = 100

x1 = 9 ; x2 = -1$

0 0

2 года назад