ai giúp em với ạ chiều nay em kiểm tra rồi. Cám ơn ạ

Question

ai giúp em với ạ chiều nay em kiểm tra rồi. Cám ơn ạ
ai-giup-em-voi-a-chieu-nay-em-kiem-tra-roi-cam-on-a

in progress 0
Phúc Điền 8 months 2021-05-10T04:38:05+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-05-10T04:39:26+00:00

    $\displaystyle\lim_{x \to 2} \dfrac{\sqrt{9-x^3}-\sqrt[3]{x^2-3}}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{\sqrt{9-x^3}-1-\sqrt[3]{x^2-3}+1}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{\sqrt{9-x^3}-1}{x^2-4}-\displaystyle\lim_{x \to 2} \dfrac{\sqrt[3]{x^2-3}-1}{x^2-4}\\ =\displaystyle\lim_{x \to 2} \dfrac{(\sqrt{9-x^3}-1)(\sqrt{9-x^3}+1)}{(x-2)(x+2)(\sqrt{9-x^3}+1)}-\displaystyle\lim_{x \to 2} \dfrac{(\sqrt[3]{x^2-3}-1)\left(\sqrt[3]{x^2-3}^2+\sqrt[3]{x^2-3}+1\right)}{(x-2)(x+2)\left(\sqrt[3]{x^2-3}^2+\sqrt[3]{x^2-3}+1\right)}\\=\displaystyle\lim_{x \to 2} \dfrac{8-x^3}{(x-2)(x+2)(\sqrt{9-x^3}+1)}-\displaystyle\lim_{x \to 2} \dfrac{x^2-4}{(x^2-4)\left(\sqrt[3]{x^2-3}^2+\sqrt[3]{x^2-3}+1\right)}\\=\displaystyle\lim_{x \to 2} \dfrac{(2-x)(4+2x+x^2}{(x-2)(x+2)(\sqrt{9-x^3}+1)}-\displaystyle\lim_{x \to 2} \dfrac{1}{\left(\sqrt[3]{x^2-3}^2+\sqrt[3]{x^2-3}+1\right)}\\=-\displaystyle\lim_{x \to 2} \dfrac{4+2x+x^2}{(x+2)(\sqrt{9-x^3}+1)}-\displaystyle\lim_{x \to 2} \dfrac{1}{\left(\sqrt[3]{x^2-3}^2+\sqrt[3]{x^2-3}+1\right)}\\=- \dfrac{4+2.2+2^2}{(2+2)(\sqrt{9-2^3}+1)}- \dfrac{1}{\left(\sqrt[3]{2^2-3}^2+\sqrt[3]{2^2-3}+1\right)}\\=-\dfrac{11}{6}$

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )