log(6) 2 * log(6) 18 + log^2(6) 3 = log(6) 2 * log(6) 6*3 + log^2(6) 3 = log(6) 2 * log(6) 3 + log(6) 2 + log^2(6) 3 = log(6) 3*(log(6) 2 + log(6) 3) + log(6) 2 = log(6) 3 *log(6) 2*3 + log(6) 2 = log(6) 2 + log(6) 3 = log(6) 2*3 = log(6) 6 = 1

log(a) b+ log(a)c = log(a) bc

0 0

4 года назад

(х-5)(2х+3)-(х+4)(4-х)=2х+5

Умербекова

2х^2+3х-10х-15-х^2-16=2х+5
2х^2+3х-10х-15-2х^2-16-2х-5=0
-5х-36=0
-5х=36
х=36/(-5)
х=-7,2
ответ 7,2

0 0

3 года назад

Проверьте, правильно ли я решил?

Niemand2 [12.2K]

$1. \ x^2 + 2x - 1 = 0\\\\
D = 4 + 4 = 8\\\\
x_1 = \frac{-2+\sqrt{8}}{2} = \frac{-2 + 2\sqrt{2}}{2} = -1 + \sqrt{2}\\\\
x_2 = \frac{-2-\sqrt{8}}{2} = \frac{-2 - 2\sqrt{2}}{2} = -1 - \sqrt{2}\\\\
(x - x_1)(x - x_2) = (x - ( -1 + \sqrt{2}))(x - (-1 - \sqrt{2})) = \\\\
= (x + 1 - \sqrt{2})(x + 1 + \sqrt{2})$

$2. \ x^2 - 2(\sqrt{2} - 1)x + 3 - 2\sqrt{2} =\\\\
= x^2 - 2(\sqrt{2} - 1)x + 2 - 2\sqrt{2} + 1 =\\\\
= x^2 - 2(\sqrt{2} - 1)x + (\sqrt{2} - 1)^2 = \\\\
= (x - (\sqrt{2} - 1))^2 = (x - \sqrt{2} + 1)^2 = (x + 1 - \sqrt{2})^2\\\\$

$3. \ \frac{x^2 + 2x - 1}{x^2 - 2(\sqrt{2} - 1)x + 3 - 2\sqrt{2}} = \frac{(x + 1 - \sqrt{2})(x + 1 + \sqrt{2})}{(x + 1 - \sqrt{2})^2} = \\\\ = \boxed{\frac{x + 1 + \sqrt{2}}{x + 1 - \sqrt{2}}}

$

0 0

3 года назад

Икс плюс 10 минус 12 равно 17

Валерий50

X+10-12=17
x+10=17-12
x+10=5
x=10+5
x=15

0 0

3 года назад

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