\(x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
e,\(\left(x^2+x+1\right)^2-2x^2-2x=5\)
Giải pt
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a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
ĐKXĐ: ...
\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)
\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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 \(\Rightarrow x=\dfrac{4\pi}{3}+k2\pi\)
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)
\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)
Kết hợp ĐKXĐ.
\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)
\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)
Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)
Ta có bảng sau:
\(x\) | \(-\infty\) \(-\dfrac{3}{2}\) \(-\dfrac{1}{2}\) \(1\) \(2\) \(+\infty\) |
\(1+2x\) | - | - 0 + | + | + |
\(x-2\) | - | - | - | - 0 + |
\(2x+3\) | - 0 + | + | + | + |
\(1-x\) | + | + | + 0 - | - |
\(f\left(x\right)\) | - || + 0 - || + 0 - |
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].
\(a.x^2+\dfrac{1}{x^2}=x+\dfrac{1}{x}\) ( ĐKXĐ : \(x\ne0\) )
\(\Leftrightarrow x^2+\dfrac{1}{x^2}-x-\dfrac{1}{x}=0\Leftrightarrow\left(x^2-\dfrac{1}{x}\right)+\left(\dfrac{1}{x^2}-x\right)=0\)
\(\Leftrightarrow-x\left(\dfrac{1}{x^2}-x\right)+\left(\dfrac{1}{x^2}-x\right)=0\Leftrightarrow\left(\dfrac{1}{x^2}-x\right)\left(1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\\dfrac{1}{x^2}-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\1-x^3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(1-x\right)\left(1+x+x^2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=1\end{matrix}\right.\Leftrightarrow x=1\) ( x2 + x + 1 loại nhé nếu phân tích ra thì ta được \(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\in R\) )
Vậy \(S=\left\{1\right\}\)
b, \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow x\left(x+3\right).\left(x+1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x+1-1\right)\left(x^2+3x+1+1\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x+1\right)-1-24=0\Leftrightarrow\left(x^2+3x+1\right)-25=0\)
\(\Leftrightarrow\left(x^2+3x+1-5\right)\left(x^2+3x+1+5\right)=0\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3x-4=0\\x^2+3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+4\right)=0\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}\forall x\in R\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{-4;1\right\}\)
e, \(\left(x^2+x+1\right)-2x^2-2x=5\Leftrightarrow\left(x^2+x+1\right)-2x^2-2x-2-3=0\)
\(\Leftrightarrow\left(x^2+x+1\right)-2\left(x^2+x+1\right)-3=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x-1\right)-3=0< =>\left(x^2+x\right)^2-4=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\) ( x^2 + x + 2 loại nhé y như mấy câu trên luôn khác 0 ! )
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;1\right\}\)