Question

Tenho muita dificuldade nessas contas
1. (3x-2y)²=
2. (3c-8d)²=
3. (9x-6y)²=
4. (10c-5d)²=
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1. x²-25=0
2. x²-49=0
3. x²-81=0
4. x²-144=0
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1. x²+5x+6=0
2. x²-5x-6=0
3. 2x²-5x+6=0
4. -x²+5x-6=0

ME AJUDE .

Answer 1

1)
$1) \\ \\ (3x - 2y)^2 \\ \\ (3x - 2y)(3x - 2y) \\ \\ 9x^2 - 6xy - 6xy + 4y^2 \\ \\ => 9x^2 - 12xy + 4y^2$

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$2) \\ \\ (3x - 8d)^2 \\ \\ (3c - 8d) (3c - 8d) \\ \\ 9c^2 - 24cd - 24cd + 64d^2 \\ \\ => 9c^2 - 48cd + 64d^2$

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$3) \\ \\ (9x - 6y)^2 \\ \\ (9x - 6y)(9x - 6y) \\ \\ 81x^2 - 54xy - 54xy + 36x^2 \\ \\ => 81x^2 - 108xy + 36x^2$

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$4) \\ \\ (10c - 5d)^2 \\ \\ (10c - 5d) (10c - 5d) \\ \\ 100c^2 - 50cd - 50cd + 25d^2 \\ \\ => 100c^2 - 100cd + 25d^2$

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$1) \\ \\ x^2-25=0 \\ \\ (x - 5)(x + 5) \\ \\ x -5 = 0 => x' = 5 \\ \\ x + 5 = 0 => x'' = -5$

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$2) \\ \\ x^2-49=0 \\ \\ (x - 7)(x + 7) \\ \\ x - 7 = 0 => x' = 7 \\ \\ x + 7 = 0 => x'' = -7$

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$3) \\ \\ x^2-81=0 \\ \\ (x - 9)(x + 9) \\ \\ x - 9 = 0 => x' = 9 \\ \\ x + 9 = 0 => x'' = -9$

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$4) \\ \\ x^2-144=0 \\ \\ (x - 12)(x + 12) \\ \\ x -12 = 0 => x' = 12 \\ \\ x + 12 = 0 => x'' = -12$

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$1) \\ \\ x^2+5x+6=0\\ \\ (x + 2)(x+3) \\ \\ x - 2 = 0 =. x' = 2 \\ \\ x + 3 = 0 => x'' = -3 \\ \\ \\ 2) \\ \\ x^2 - 5x - 6 = 0 \\ \\ (x - 6)(x + 1) \\ \\ x - 6 = 0 => x' = 6 \\ \\ x + 1 = 0 => x'' = -1 \\ \\ \\ 3) \\ \\ 2x^2 - 5x + 6 = 0$

3) sem solução para os números reais:$4) \\ \\ -x^2 + 5x - 6 = 0 \\ \\ -((x - 3)(x - 2)) = > (x - 3)(x - 2) \\ \\ x - 3 = 0 => x' = 3 \\ \\ x - 2 = 0 => x'' = 2$ Δ = -23