Đề 1:Bài 1: Giải và biện luận các phương trình sau( gồm cả phương trình lượng giác):1) $A^{3}_{2x-1}+A^{4}_{2x-2}=3x^2+P_{2}$2) $ 2\sin(6x+\frac{\pi}{3})\cos(2x)=\tan(3x-\frac{\pi}{4})\tan(3x-\frac{\pi}{6})$3)...
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$\mathop {\lim }\limits_{x \to 1}\frac{5}{(x-1)(x^2-3x+2)}$
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Tính : $\mathop {\lim }\limits_{x \to 1}\frac{x^{2016}-2016x+2015}{(x-1)^2}$
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\[\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 + x} .\sqrt[3]{{1 + \frac{x}{2}}}.\sqrt[4]{{1 + \frac{x}{3}}} - \sqrt[4]{{1 - x}}}}{{\frac{3}{2}\sqrt {4 + x} - \sqrt[3]{{3 - x}} - \sqrt[4]{{1 + x}}}}\]
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$g(x)=\left\{ \begin{array}{l} \frac{x^3-8}{\sqrt{x+2}-2} ,khi x > 2\\ 20x+8 ,khi x <2\\ m^2-5m+52, khi x = 2 \end{array} \right.$ Tại $x_0 = 2$
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$L=\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{cosx}-\sqrt[3]{cosx}}{sin^2x}$
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$\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x^2-2x+1}$
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$\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x^2-2x+1}$
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1) Tìm giới hạn của hàm số sau:$\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[3]{6x+1}-\sqrt{4x+1}}{1-cos2x}.$
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a)$\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}$ b)$\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}$c)$\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$
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a)$\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}$ b)$\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}$c)$\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$
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a)$\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1}$ b)$\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}$c)$\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}$
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tìm các giới hạn sau.$\mathop {\lim \frac{1-cos6x}{x^{2}cosx} }\limits_{x \to 0}$$\mathop {\lim \frac{1+sinx-cosx}{1-sinx-cosx}}\limits_{x \to 0}$
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1) Tính các giới hạn sau:$a/\mathop {\lim }\limits_{x \to 2}\frac{(2x-1)^5-243}{x-2} b/\mathop {\lim }\limits_{x \to 4}\frac{\sqrt{(x-3)^3}-1}{16(2x-7)^5-x^2}$*giải kĩ hộ em với, có giải nhưng khó hiêu quá!
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1) Tính các giới hạn sau:$a/\mathop {\lim }\limits_{x \to 2}\frac{(2x-1)^5-243}{x-2} b/\mathop {\lim }\limits_{x \to 4}\frac{\sqrt{(x-3)^3}-1}{16(2x-7)^5-x^2}$*giải kĩ hộ em với, có giải nhưng khó hiêu quá!
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1) $\mathop {\lim }\limits_{x \to 1}\frac{x^m-1}{x^n-1}$2) $\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x-1}.$
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Tìm $\mathop {\lim }\limits_{x \to 1} \frac{x^{100}-1}{x^{99}-1}$
Trả lời 20-03-14 08:34 PM
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4) $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+x}\sqrt[3]{1+\frac{x}{2}}\sqrt[4]{1+\frac{x}{3}}-\sqrt[4]{1-x}}{\frac{3}{2}\sqrt{4+x}-\sqrt[3]{8-x}-\sqrt[4]{1-x}}$5)$\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[3]{6x^2+2}-2\sqrt{x}}{x^3-x^2-x+1}$
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4) $\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+x}\sqrt[3]{1+\frac{x}{2}}\sqrt[4]{1+\frac{x}{3}}-\sqrt[4]{1-x}}{\frac{3}{2}\sqrt{4+x}-\sqrt[3]{8-x}-\sqrt[4]{1-x}}$5)$\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[3]{6x^2+2}-2\sqrt{x}}{x^3-x^2-x+1}$
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1) $\mathop {\lim }\limits_{x \to 0} \frac{(1+x)(1+2x)(1+3x)(1+4x)-1}{x}$2)$\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[4]{1+4x}-1}{x}$3)$\mathop {\lim }\limits_{x \to 0}\frac{(x^2+2014)\sqrt[7]{1-2x}-2014}{x}$cách làm bài...
Trả lời 17-03-14 07:07 AM
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