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b/ \(\sqrt{12-\dfrac{12}{x^2}}+\sqrt{x^2-\dfrac{12}{x^2}}=x^2\)
\(\Leftrightarrow x-\sqrt{12-\dfrac{12}{x^2}}=\sqrt{x^2-\dfrac{12}{x^2}}\)
Bình phương 2 vế rút gọn
\(\Leftrightarrow x^4-x^2-4\sqrt{3\left(x^4-x^2\right)}+12=0\)
Đặt \(\sqrt{x^4-x^2}=a\)
\(\Rightarrow a^2-4\sqrt{3}a+12=0\)
\(\Leftrightarrow a=2\sqrt{3}\)
\(\Leftrightarrow x^4-x^2=12\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
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\(1.\left(x\ne\pm1\right)\Rightarrow pt\Leftrightarrow\left(x-m\right)\left(x-1\right)=\left(x+1\right)\left(x-2\right)\)
\(\Leftrightarrow x^2-x\left(m+1\right)+m=x^2-x-2\)
\(\Leftrightarrow-x\left(m+1\right)+m=-x-2\)
\(\Leftrightarrow x=\dfrac{m+2}{m}\left(m\ne0\right)\)
\(pt-có-ngo-duy-nhất\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m+2}{m}\ne1\\\dfrac{m+2}{m}\ne-1\end{matrix}\right.\)\(\Leftrightarrow m\ne-1\)
\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\m\ne-1\end{matrix}\right.\)
\(2.\left\{{}\begin{matrix}x^2+8y^2=12\left(1\right)\\x^3+2xy^2+12y=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x^3+2xy^2+y\left(x^2+8y^2\right)=0\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2y\left(3\right)\\x^2-xy+4y^2=\left(x-\dfrac{y}{2}\right)^2+\dfrac{15}{4}y^2=0\left(4\right)\end{matrix}\right.\)
\(\left(3\right)\left(1\right)\Rightarrow4y^2+8y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)
với \(x=y=0\) không là nghiệm của hệ pt
với \(x=y\ne0\Rightarrow\left(4\right)>0\Rightarrow\left(4\right)-vô-nghiệm\)
\(\Rightarrow\left(x;y\right)=\left\{\left(-2;1\right);\left(2;-1\right)\right\}\)
\(1,\Leftrightarrow\left(x-m\right)\left(x-1\right)=x^2-x-2\\ \Leftrightarrow x^2-x-mx+m-x^2+x+2=0\\ \Leftrightarrow mx=m+2\)
PT có nghiệm duy nhất \(\Leftrightarrow m\ne0\)
\(2,\Leftrightarrow\left\{{}\begin{matrix}x^2y+8y^3=12y\\x^3+2xy^2+12y=0\end{matrix}\right.\)
Thế \(PT\left(1\right)\rightarrow PT\left(2\right)\Leftrightarrow x^3+2xy^2+x^2y+8y^3=0\)
\(\Leftrightarrow\left(x+2y\right)\left(x^2-2xy+4y^2\right)+xy\left(x+2y\right)=0\\ \Leftrightarrow\left(x+2y\right)\left(x^2-xy+4y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left(x-\dfrac{1}{2}y\right)^2+\dfrac{15}{4}y^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\\left\{{}\begin{matrix}x-\dfrac{1}{2}y=0\\y^2=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2y\\x=y=0\end{matrix}\right.\)
Thay \(x=y=0\Leftrightarrow0+0=12\left(loại\right)\)
Thay \(x=-2y\Leftrightarrow4y^2+8y^2=12y^2=12\Leftrightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-2\\y=-1\Rightarrow x=2\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\left(-2;1\right);\left(2;-1\right)\right\}\)
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Ta có:
$p^2=5q^2+4$ chia 5 dư 4 suy ra $p=5k+2(k\in \mathbb{N}^*)$
Ta có:
$(5k+2)^2=5q^2+4\Leftrightarrow 5k^2+4k=q^2\Rightarrow q^2\vdots k$
Mặt khác q là số nguyên tố và $q>k$ nên $k=1$. Thay vào ta được $p=7,q=3$
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https://diendantoanhoc.net/topic/163051-x-fracxsqrtx2-1-frac3512/
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a.
\(\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=3sinx+cosx+2\)
\(\Leftrightarrow sin2x+cos2x=3sinx+cosx+2\)
\(\Leftrightarrow2sinx.cosx-3sinx+2cos^2x-cosx-3=0\)
\(\Leftrightarrow sinx\left(2cosx-3\right)+\left(cosx+1\right)\left(2cosx-3\right)=0\)
\(\Leftrightarrow\left(2cosx-3\right)\left(sinx+cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{3}{2}\left(vn\right)\\sinx+cosx+1=0\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(cosx\ne\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x\ne\dfrac{\pi}{3}+k2\pi\\x\ne-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\dfrac{\left(2-\sqrt{3}\right)cosx-2sin^2\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)}{2cosx-1}=1\)
\(\Rightarrow\left(2-\sqrt{3}\right)cosx+cos\left(x-\dfrac{\pi}{2}\right)=2cosx\)
\(\Leftrightarrow-\sqrt{3}cosx+sinx=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=0\)
\(\Rightarrow x-\dfrac{\pi}{3}=k\pi\)
\(\Rightarrow x=\dfrac{\pi}{3}+k\pi\)
Kết hợp ĐKXĐ \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\)
- Với \(x< -1\Rightarrow x+\dfrac{x}{\sqrt{x^2-1}}< 0\) pt vô nghiệm
- Xét với \(x>1\):
Bình phương 2 vế của pt đã cho:
\(x^2+\dfrac{x^2}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}=\dfrac{1225}{144}\)
\(\Leftrightarrow\dfrac{x^4}{x^2-1}+\dfrac{2x^2}{\sqrt{x^2-1}}-\dfrac{1225}{144}=0\)
Đặt \(\dfrac{x^2}{\sqrt{x^2-1}}=t>0\)
\(\Rightarrow t^2+2t-\dfrac{1225}{144}=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{25}{12}\\t=-\dfrac{49}{12}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{x^2}{\sqrt{x^2-1}}=\dfrac{25}{12}\)
Tới đây có thể bình phương 2 vế hoặc đặt \(\sqrt{x^2-1}=a\Rightarrow x^2=a^2+1\) đưa về pt bậc 2:
\(\dfrac{a^2+1}{a}=\dfrac{25}{12}\Leftrightarrow a^2-\dfrac{25}{12}a+1=0\) \(\Rightarrow a=...\Rightarrow x=...\)