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PT 2
\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))
\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)
\(\Rightarrow2x^2-3x+6=0\)
=> PT vô nghiệm.
\(ĐK:x\ne3;x\ne2\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1+\dfrac{x^2}{x-2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\\dfrac{x^2+x+2}{x-2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x^2+x+2=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
ĐKXĐ: \(x\notin\left\{2;-1;\dfrac{-3\pm\sqrt{17}}{2}\right\}\)
\(\dfrac{x}{x^2-x-2}+\dfrac{3x}{x^2+3x-2}=1\)
=>\(\dfrac{x\left(x^2+3x-2\right)+3x\left(x^2-x-2\right)}{\left(x^2-x-2\right)\left(x^2+3x-2\right)}=1\)
=>\(\dfrac{x^3+3x^2-2x+3x^3-3x^2-6x}{\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2}=1\)
=>\(4x^3-8x=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)
=>\(4x\left(x^2-2\right)=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)
=>\(\left(x^2-2\right)^2-2x\left(x^2-2\right)-3x^2=0\)
=>\(\left(x^2-2\right)^2-3x\left(x^2-2\right)+x\left(x^2-2\right)-3x^2=0\)
=>\(\left(x^2-2\right)\left(x^2-2-3x\right)+x\left(x^2-2-3x\right)=0\)
=>\(\left(x^2+x-2\right)\left(x^2-3x-2\right)=0\)
=>\(\left(x+2\right)\left(x-1\right)\left(x^2-3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=1\left(nhận\right)\\x=\dfrac{3\pm\sqrt{17}}{2}\left(nhận\right)\end{matrix}\right.\)
ĐKXĐ : \(x\inℝ\)
Ta có : \(\dfrac{x^2+4x+5}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)
\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)
Phương trình (1) <=> 5(x2 - 3x + 5) = 3(x2 - x + 5)
<=> 2x2 - 12x + 10 = 0
<=> x2 - 6x + 5 = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Tập nghiệm \(S=\left\{0;1;5\right\}\)
2: ĐKXĐ: x>=0
\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)
=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)
=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)
=>\(-2\sqrt{3x}=-4\)
=>\(\sqrt{3x}=2\)
=>3x=4
=>\(x=\dfrac{4}{3}\left(nhận\right)\)
3:
ĐKXĐ: x>=0
\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)
=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)
=>\(13\sqrt{2x}=20+3\sqrt{2}\)
=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)
=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)
=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)
4: ĐKXĐ: x>=-1
\(\sqrt{16x+16}-\sqrt{9x+9}=1\)
=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>\(\sqrt{x+1}=1\)
=>x+1=1
=>x=0(nhận)
5: ĐKXĐ: x<=1/3
\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)
=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)
=>\(5\sqrt{1-3x}=10\)
=>\(\sqrt{1-3x}=2\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1(nhận)
6: ĐKXĐ: x>=3
\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\left(\dfrac{2}{3}+\dfrac{1}{6}-1\right)=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)
=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)
=>x-3=16
=>x=19(nhận)
Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)
\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)
Lấy (2)-(1) ta đc:
\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(x-1\right)\\1=y+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Với \(x\ne1;y\ne-2\)
hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}2x-2=x\\\dfrac{1}{y+2}=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=2\\y+2=1\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
\(ĐK:x\ne3\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(-x-1+x^2-2x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\x^2-3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=\dfrac{3+\sqrt{41}}{2}\\x=\dfrac{3-\sqrt{41}}{2}\end{matrix}\right.\)
