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giúp em bài này với em cảm ơn ạ

Hồng Phúc · Accepted Answer

$L=\lim\limits_{x\rightarrow+\infty}\left(2x^2-\sqrt{x^2-x}.\sqrt[3]{8x^3+12x^2-3x}\right)$Đặt $f\left(x\right)=2x^2-\sqrt{x^2-x}.\sqrt[3]{8x^3+12x^2-3x}$Ta có:$2.f\left(x\right)=4x^2-\sqrt{4x^2-4x}.\sqrt[3]{8x^3+12x^2-3x}$$=1+\left(4x^2-1\right)-\sqrt{4x^2-4x}.\sqrt[3]{8x^3+12x^2-3x}$$=1+\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)+\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$Đặt $A\left(x\right)=\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)$$B\left(x\right)=\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$$A\left(x\right)=\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)$$=\dfrac{\left(2x-1\right)\left(8x^3+12x^2+6x+1-8x^3-12x^2+3x\right)}{\left(2x+1\right)^2+\sqrt[3]{\left(8x^3+12x^2-3x\right)^2}+\left(2x+1\right)\sqrt[3]{8x^3+12x^2-3x}}$$=\dfrac{\left(2x-1\right)\left(9x+1\right)}{\left(2x+1\right)^2+\sqrt[3]{\left(8x^3+12x^2-3x\right)^2}+\left(2x+1\right)\sqrt[3]{8x^3+12x^2-3x}}$$\Rightarrow\lim\limits_{x\rightarrow+\infty}A\left(x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2-\dfrac{1}{x}\right)\left(9+\dfrac{1}{x}\right)}{\left(2+\dfrac{1}{x}\right)^2+\sqrt[3]{\left(8+\dfrac{12}{x}-\dfrac{3}{x^2}\right)^2}+\left(2+\dfrac{1}{x}\right)\sqrt[3]{8+\dfrac{12}{x}-\dfrac{3}{x^2}}}$$=\dfrac{2.9}{2^2+4+2.2}$$=\dfrac{3}{2}$$B\left(x\right)=\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$$=\dfrac{\left(4x^2-4x+1-4x^2+4x\right).\sqrt[3]{8x^3+12x^2-3x}}{2x-1+\sqrt{4x^2-4x}}$$=\dfrac{\sqrt[3]{8x^3+12x^2-3x}}{2x-1+\sqrt{4x^2-4x}}$$\Rightarrow\lim\limits_{x\rightarrow+\infty}B\left(x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt[3]{8+\dfrac{12}{x}-\dfrac{3}{x^2}}}{2-\dfrac{1}{x}+\sqrt{4-\dfrac{4}{x}}}$$=\dfrac{2}{2+2}$$=\dfrac{1}{2}$$\Rightarrow2L=\lim\limits_{x\rightarrow+\infty}\left[2f\left(x\right)\right]$$=\lim\limits_{x\rightarrow+\infty}\left[1+A\left(x\right)+B\left(x\right)\right]$$=1+\lim\limits_{x\rightarrow+\infty}A\left(x\right)+\lim\limits_{x\rightarrow+\infty}B\left(x\right)$$=1+\dfrac{3}{2}+\dfrac{1}{2}$$=3$$\Rightarrow L=\dfrac{3}{2}$Câu trả lời: $L=\lim\limits_{x\rightarrow+\infty}\left(2x^2-\sqrt{x^2-x}.\sqrt[3]{8x^3+12x^2-3x}\right)$Đặt $f\left(x\right)=2x^2-\sqrt{x^2-x}.\sqrt[3]{8x^3+12x^2-3x}$Ta có:$2.f\left(x\right)=4x^2-\sqrt{4x^2-4x}.\sqrt[3]{8x^3+12x^2-3x}$$=1+\left(4x^2-1\right)-\sqrt{4x^2-4x}.\sqrt[3]{8x^3+12x^2-3x}$$=1+\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)+\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$Đặt $A\left(x\right)=\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)$$B\left(x\right)=\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$$A\left(x\right)=\left(2x-1\right)\left(2x+1-\sqrt[3]{8x^3+12x^2-3x}\right)$$=\dfrac{\left(2x-1\right)\left(8x^3+12x^2+6x+1-8x^3-12x^2+3x\right)}{\left(2x+1\right)^2+\sqrt[3]{\left(8x^3+12x^2-3x\right)^2}+\left(2x+1\right)\sqrt[3]{8x^3+12x^2-3x}}$$=\dfrac{\left(2x-1\right)\left(9x+1\right)}{\left(2x+1\right)^2+\sqrt[3]{\left(8x^3+12x^2-3x\right)^2}+\left(2x+1\right)\sqrt[3]{8x^3+12x^2-3x}}$$\Rightarrow\lim\limits_{x\rightarrow+\infty}A\left(x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\left(2-\dfrac{1}{x}\right)\left(9+\dfrac{1}{x}\right)}{\left(2+\dfrac{1}{x}\right)^2+\sqrt[3]{\left(8+\dfrac{12}{x}-\dfrac{3}{x^2}\right)^2}+\left(2+\dfrac{1}{x}\right)\sqrt[3]{8+\dfrac{12}{x}-\dfrac{3}{x^2}}}$$=\dfrac{2.9}{2^2+4+2.2}$$=\dfrac{3}{2}$$B\left(x\right)=\left(2x-1-\sqrt{4x^2-4x}\right).\sqrt[3]{8x^3+12x^2-3x}$$=\dfrac{\left(4x^2-4x+1-4x^2+4x\right).\sqrt[3]{8x^3+12x^2-3x}}{2x-1+\sqrt{4x^2-4x}}$$=\dfrac{\sqrt[3]{8x^3+12x^2-3x}}{2x-1+\sqrt{4x^2-4x}}$$\Rightarrow\lim\limits_{x\rightarrow+\infty}B\left(x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt[3]{8+\dfrac{12}{x}-\dfrac{3}{x^2}}}{2-\dfrac{1}{x}+\sqrt{4-\dfrac{4}{x}}}$$=\dfrac{2}{2+2}$$=\dfrac{1}{2}$$\Rightarrow2L=\lim\limits_{x\rightarrow+\infty}\left[2f\left(x\right)\right]$$=\lim\limits_{x\rightarrow+\infty}\left[1+A\left(x\right)+B\left(x\right)\right]$$=1+\lim\limits_{x\rightarrow+\infty}A\left(x\right)+\lim\limits_{x\rightarrow+\infty}B\left(x\right)$$=1+\dfrac{3}{2}+\dfrac{1}{2}$$=3$$\Rightarrow L=\dfrac{3}{2}$

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