Jak obliczyć mnożenie sum algebraicznych: (2√3 - 3√2)(√2 + √3) - (4 - √6) (√5+2√3)(2√5 - √3) + (6

January 2019 1 32 Report

Jak obliczyć mnożenie sum algebraicznych:
(2√3 - 3√2)(√2 + √3) - (4 - √6)
(√5+2√3)(2√5 - √3) + (6 - 3√15)

-5x² + 3x(x - 1) + (2x- 1)(x + 3)
3y² - 2x(x + 2y) - (x - y)(2x + y)
-3x(x -1)(x-2) - 9x²
(a + 2c)(a - c + 2) - ac
Proszę o wytłumaczenie :).

ghe $(2 \sqrt{3} - 3 \sqrt{2} )( \sqrt{2} + \sqrt{3} ) - (4 - \sqrt{6} )=$

$2 \sqrt{3} \cdot \sqrt{2}+2 \sqrt{3} \cdot \sqrt{3}-3 \sqrt{2} \cdot \sqrt{2} -3 \sqrt{2}\cdot \sqrt{3}-4 + \sqrt{6} =$

$2 \sqrt{6} +6-6 -3 \sqrt{6}-4 + \sqrt{6} =$

$-4$
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$( \sqrt{5} +2 \sqrt{3} )(2 \sqrt{5} - \sqrt{3}) + (6 - 3 \sqrt{15})=$

$\sqrt{5} \cdot 2 \sqrt{5} -\sqrt{5} \cdot \sqrt{3}+2 \sqrt{3} \cdot 2 \sqrt{5} - 2 \sqrt{3} \cdot \sqrt{3} + 6 - 3 \sqrt{15}=$

$10-\sqrt{15} +4 \sqrt{15} - 6+ 6 - 3 \sqrt{15}=$

$10$
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$-5x^2 + 3x(x - 1) + (2x- 1)(x + 3)=$

$-5x^2 + 3x^2-3x + 2x^2+6x-x-3=$

$2x-3$
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$3y^2 - 2x(x + 2y) - (x - y)(2x + y)=$

$3y^2 - 2x^2-4xy - (2x^2+xy-2xy-y^2)=$

$3y^2 - 2x^2-4xy - 2x^2-xy+2xy+y^2=$

$3y^2 - 2x^2-4xy - 2x^2-xy+2xy+y^2=$

$- 4x^2 - 3xy + 4y^2$
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$-3x(x -1)(x-2) - 9x^2=$

$-3x(x^2-2x-x+2) - 9x^2=$

$-3x^3+6x^2+3x^2-6x - 9x^2=$

$- 3x^3 - 6x$
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$(a + 2c)(a - c + 2) - ac =$

$a^2-ac+2a+2ac-2c^2+4c - ac =$

$a^2 + 2a - 2c^2 + 4c$

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