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trlong12345

trlong12345

Đăng ký: 01-05-2012
Offline Đăng nhập: 20-06-2014 - 09:18
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#318655 $3cotx-tanx=8sin(x-\frac{8\pi}{3})$

Gửi bởi trlong12345 trong 23-05-2012 - 09:02

$3cotx-tanx=8sin(x-\frac{8\pi}{3})$

ĐK $x\neq k\frac{\pi}{2}$
$3cotx-tanx=8sin(x-\frac{8\pi}{3})$
$\Leftrightarrow$$3\frac{cosx}{sinx}-\frac{sinx}{cosx}=8sin(\frac{\pi}{3}-x)$
$\Leftrightarrow$$\frac{3cos^{2}x-sin^{2}x}{sinxcosx}=8(\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx)$
$\Leftrightarrow$$(\sqrt{3}cosx-sinx)(\sqrt{3}cosx+sinx)-4sinxcosx(\sqrt{3}cosx-sinx)=0$
$\Leftrightarrow$$(\sqrt{3}cosx-sinx)(\sqrt{3}cosx+sinx-4sinxcosx)=0$
$\Leftrightarrow$$(\sqrt{3}cosx-sinx)(2sin(x+\frac{\pi}{3})-2sin2x)=0$
$\Leftrightarrow$$\left[\begin{array}{1} \sqrt{3}cosx-sinx=0 \\ sin(x+\frac{\pi}{3})=sin2x \end{array}\right.$
$\Leftrightarrow$$\left[\begin{array}{1} x=\frac{\pi}{3}+k\pi \\ x=\frac{2\pi}{9}+k\frac{2\pi}{3} \end{array}\right.$ ($k\in Z$)


#318648 $\frac{2cos^{2}x+2cosx-3}{sin^{2}\frac{x}{2}}+4\sqrt{3}si...

Gửi bởi trlong12345 trong 23-05-2012 - 08:18

$\frac{2cos^{2}x+2cosx-3}{sin^{2}\frac{x}{2}}+4\sqrt{3}sinx=0$

DK $x\neq k2\pi$
$\frac{2cos^{2}x+2cosx-3}{sin^{2}\frac{x}{2}}+4\sqrt{3}sinx=0$
$\Leftrightarrow$$\frac{4cos^{2}x+4cosx-6}{1-cosx}+4\sqrt{3}sinx=0$
$\Leftrightarrow$$4cos^{2}x+4cosx-6+4\sqrt{3}sinx(1-cosx)=0$
$\Leftrightarrow$$-2cos^{2}x+4cosx-6sin^{2}x+4\sqrt{3}sinx(1-cosx)=0$
$\Leftrightarrow$$(-2cos^{2}x-2\sqrt{3}sinxcosx)+(4cosx+4\sqrt{3}sinx)+(-6sin^{2}x-2\sqrt{3}sinxcosx)=0$
$\Leftrightarrow$$-2cosx(cosx+\sqrt{3}sinx)+4(cosx+\sqrt{3}sinx)-2\sqrt{3}sinx(cosx+\sqrt{3}sinx)=0$
$\Leftrightarrow$$(cosx+\sqrt{3}sinx)(2cosx+2\sqrt{3}sinx-4)=0$
Đến đây là dc rùi :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2: :ukliam2:


#314818 1.Tính: $\lim_{x\to1}\frac{x^{n}-nx+n-1}{(x-1)^{2}}$

Gửi bởi trlong12345 trong 07-05-2012 - 07:48

1.Tính đạo hàm : $\lim_{x\to1}\frac{x^{n}-nx+n-1}{(x-1)^{2}}$

mình chỉ bít tính giới hạn ko hiểu tính đạo hàm cái này thế nào :icon6:
$\lim_{x\to1}$$\frac{x^{n}-nx+n-1}{(x-1^{2})}$
=$\lim_{x\to1}$$\frac{(x^{n}-1)-n(x-1)}{(x-1)^{2}}$
=$\lim_{x\to1}$$\frac{x^{n-1}+x^{n-2}+...+1-n}{x-1}$
=$\lim_{x\to1}$$\frac{(x^{n-1}-1)+(x^{n-2}-1)+...+(x-1)}{x-1}$
=$\lim_{x\to1}$$(x^{n-2}+x^{n-3}+...+1)+(x^{n-3}+x^{n-4}+...+1)+...+(x+1)+1$
=$(n-1)+(n-2)+...+2+1$=$\frac{n(n-1)}{2}$


#313764 Tính giới hạn sau :$\lim_{x\to a} \frac{\sqrt[3]{cos...

Gửi bởi trlong12345 trong 01-05-2012 - 20:45

$\lim_{x \to a}$$\frac{\sqrt[3]{cosx} -\sqrt[3]{cosa}}{x^{2} -a^{2}}$
=$\lim_{x \to a}$$\frac{cosx -cosa}{(x^{2}-a^{2})(\sqrt[3]{cos^{2}x}+\sqrt[3]{cosxcosa}+\sqrt[3]{cos^{2}}a)}$
=$\lim_{x \to a}$$\frac{\frac{2sin^{2}\frac{a}{2}}{(\frac{a}{2})^{2}}(\frac{a}{2})^{2}-\frac{2sin^{2}\frac{x}{2}}{(\frac{x}{2})^{2}}(\frac{x}{2})^{2}}{(x^{2}-a^{2})(\sqrt[3]{cos^{2}x}+\sqrt[3]{cosxcosa}+\sqrt[3]{cos^{2}a})}$
=$\lim_{x \to a}$$\frac{2(\frac{a}{2})^{2}-2(\frac{x}{2})^{2}}{(x^{2}-a^{2})(\sqrt[3]{cos^{2}x}+\sqrt[3]{cosxcosa}+\sqrt[3]{cos^{2}a})}$
=$\lim_{x \to a}$$\frac{-1}{6\sqrt[3]{cos^{2}a}}$