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a) Đặt \(t=\left|2x-\dfrac{1}{x}\right|\Leftrightarrow t^2=\left(2x-\dfrac{1}{x}\right)^2=4x^2-4+\dfrac{1}{x^2}\Leftrightarrow t^2+4=4x^2+\dfrac{1}{x^2}\) ĐK \(t\ge0\)
từ có ta có pt theo biến t : \(t^2+4+t-6=0\)
\(\Leftrightarrow t^2+t-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\left|2x-\dfrac{1}{x}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{x}=1\\2x-\dfrac{1}{x}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x^2-x-1=0\\2x^2+x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
c: TH1: x>0
Pt sẽ là \(\dfrac{x^2-1}{x\left(x-2\right)}=2\)
=>2x^2-4x=x^2-1
=>x^2-4x+1=0
hay \(x=2\pm\sqrt{3}\)
TH2: x<0
Pt sẽ là \(\dfrac{x^2-1}{-x\left(x-2\right)}=2\)
=>-2x(x-2)=x^2-1
=>-2x^2+4x=x^2-1
=>-3x^2+4x+1=0
hay \(x=\dfrac{2-\sqrt{7}}{3}\)
b:
TH1: 2x^3-x>=0
\(4x^4+6x^2\left(2x^3-x\right)+1=0\)
=>4x^4+12x^5-6x^3+1=0
\(\Leftrightarrow x\simeq-0.95\left(loại\right)\)
TH2: 2x^3-x<0
Pt sẽ là \(4x^4+6x^2\left(x-2x^3\right)+1=0\)
=>4x^4+6x^3-12x^5+1=0
=>x=0,95(loại)
a,\(\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)
<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{2x-1}{2}=1-\dfrac{x^2-x-3}{1-x}\)
<=>\(\dfrac{5x-2}{2\left(1-x\right)}+\dfrac{\left(2x-1\right)\left(1-x\right)}{2\left(1-x\right)}=\dfrac{2\left(1-x\right)}{2\left(1-x\right)}-\dfrac{2\left(x^2-x-3\right)}{2\left(1-x\right)}\)
=>\(5x-2+2x-2x^2-1+x=2-2x-2x^2+2x+6\)
<=>\(-2x^2+8x-3=-2x^2+8\)
<=>\(8x=11< =>x=\dfrac{11}{8}\)
vậy..........
b,\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)
<=>\(\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(3x-1\right)+1}{\left(x-2\right)\left(x+2\right)}\)
=>\(x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-x+1\)
<=>\(3x^2-25x-6=3x^2-x+1\)
<=>\(-24x=7< =>x=\dfrac{-7}{24}\)
vậy..................
câu c tương tự nhé :)
a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\\dfrac{1}{x}=-3+\dfrac{4}{y}=-3+4=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)
1. \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=5\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=9\end{matrix}\right.\) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\y>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y+xy^2+x+y=5xy\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^4y^2+x^2y^4+x^2+y^2=25x^2y^2\\x^4y^2+x^2y^4+x^2+y^2=9x^2y^2\end{matrix}\right.\)\(\Leftrightarrow0=16x^2y^2\)
\(\Rightarrow\) phương trình vô nghiệm
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
điều kiện \(\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\)
Đặt \(\dfrac{x-1}{x+2}=p\left(p\ne0\right)\) và \(\dfrac{x-3}{x+2}=q\), khi đó pt đã cho trở thành \(p^2+q-2.\left(\dfrac{q}{p}\right)^2=0\) (vì \(\dfrac{q}{p}=\dfrac{\dfrac{x-3}{x+2}}{\dfrac{x-1}{x+2}}=\dfrac{x-3}{x-1}\))
\(\Leftrightarrow p^2+q-\dfrac{2q^2}{p^2}=0\)
\(\Leftrightarrow p^4+p^2q-2q^2=0\) (do \(p\ne0\) nên ta có thể chia cả 2 vế của pt cho \(p\))
\(\Leftrightarrow p^4-p^2q+2p^2q-2q^2=0\)
\(\Leftrightarrow p^2\left(p^2-q\right)+2q\left(p^2-q\right)=0\)
\(\Leftrightarrow\left(p^2-q\right)\left(p^2+2q\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}p^2-q=0\\p^2+2q=0\end{matrix}\right.\)
Nếu \(p^2-q=0\) thì \(\left(\dfrac{x-1}{x+2}\right)^2-\dfrac{x-3}{x+2}=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{\left(x+2\right)^2}-\dfrac{\left(x-3\right)\left(x+2\right)}{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1-x^2-2x+3x+6}{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\dfrac{-x+7}{\left(x+2\right)^2}=0\)
\(\Rightarrow x-7=0\)
\(\Leftrightarrow x=7\left(nhận\right)\)
Nếu \(p^2+2q=0\) thì \(\left(\dfrac{x-1}{x+2}\right)^2+2.\left(\dfrac{x-3}{x+2}\right)=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1}{\left(x+2\right)^2}+\dfrac{2\left(x-3\right)\left(x+2\right)}{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\dfrac{x^2-2x+1+2x^2+4x-6x-12}{\left(x+2\right)^2}=0\)
\(\Leftrightarrow\dfrac{3x^2-4x-11}{\left(x+2\right)^2}=0\)
\(\Rightarrow3x^2-4x-11=0\) (*)
Ta thấy \(\Delta'=\left(-2\right)^2-3.\left(-11\right)=37>0\)
Do đó pt (*) có 2 nghiệm phân biệt:
\(x_1=\dfrac{2+\sqrt{37}}{3}\left(nhận\right)\)
\(x_2=\dfrac{2-\sqrt{37}}{3}\left(nhận\right)\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{7;\dfrac{2\pm\sqrt{37}}{3}\right\}\)