<span><span>![(2a-3)^{2}-4a(a+1)=4a^{2}-12a+9-4a^{2}-4a=(4a^{2}-4a^{2})+(-12a-4a)+9=-16a+9=9-16a](https://tex.z-dn.net/?f=%282a-3%29%5E%7B2%7D-4a%28a%2B1%29%3D4a%5E%7B2%7D-12a%2B9-4a%5E%7B2%7D-4a%3D%284a%5E%7B2%7D-4a%5E%7B2%7D%29%2B%28-12a-4a%29%2B9%3D-16a%2B9%3D9-16a)
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Решение
<span>Найдите угол между векторами а(– 1; – 1) и в(2; 0) .
cos(a</span>∧b) = (x₁ *x₂ + y₁ * y₂) / [IaI*IbI]
IaI = √[(-1)² + (-1)²] = √2
IbI = √(2² + 0²) = √4 = 2
cos(a∧b) = [(-1)*2 + (-1)*0] / (2√2) = - 2/(2√2) = - 1/√2
(a<span>∧b) = 3</span>π/4