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\(\Rightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+10x+8x+80}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{x\left(x+1\right)+3\left(x+1\right)}+\frac{5}{x\left(x+3\right)+8\left(x+3\right)}+\frac{2}{x\left(x+10\right)+8\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\Rightarrow\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\Rightarrow\frac{9}{x^2+11x+10}=\frac{9}{52}\)
\(\Rightarrow x^2+11x+10=52\Rightarrow x^2+2\cdot\frac{11}{2}x+\frac{121}{4}-\frac{81}{4}=52\)
\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{289}{4}\Rightarrow x+\frac{11}{2}=\frac{17}{2}\Rightarrow x=\frac{17}{2}-\frac{11}{2}=\frac{6}{2}=3\Rightarrow x=3\)
\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)(ĐKXĐ: x khác -1;-3;-8;-10)
\(\Leftrightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+8x+10x+80}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2\left(x+8\right)\left(x+10\right)+5\left(x+1\right)\left(x+10\right)+2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9x^2+99x+216}{x^4+22x^3+155x^2+374x+240}=\frac{9}{52}\)
\(\Rightarrow468x^2+5148x+11232=9x^4+198x^3+1395x^2+3366x+2160\)
\(\Leftrightarrow9x^4+198x^3+927x^2-1782x-9072=0\)
\(\Leftrightarrow x^4+22x^3+103x^2-198x-1008=0\)
\(\Leftrightarrow x^4-3x^3+25x^3-75x^2+178x^2-534x+336x-1008=0\)
\(\Leftrightarrow x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+25x^2+178x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2+22x^2+66x+112x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+22x\left(x+3\right)+112\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+22x+112\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+8x+14x+112\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left[x\left(x+8\right)+14\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+8\right)\left(x+14\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}}{\orbr{\begin{cases}x+8=0\\x+14=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-3\left(\times\right)\\x=3\end{cases}}}{\orbr{\begin{cases}x=-8\left(\times\right)\\x=-14\end{cases}}}\)(Vì x=-3 và x=-8 không t/m ĐKXĐ)
Vậy tập nghiệm của pt là \(S=\left\{3;-14\right\}.\)
b: \(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-1\right)\left(x+2\right)}=\dfrac{-4x^2+11x-2}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x^2+4x+4+4x^2-11x+2=0\)
\(\Leftrightarrow5x^2-7x+6=0\)
hay \(x\in\varnothing\)
c: \(\Leftrightarrow\left(3x^2+2\right)^2-5x\left(3x^2+2\right)=0\)
=>3x^2-5x+2=0
=>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
2. ĐK: \(x\ge0\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x}\ge0\\b=\sqrt{x^2+4}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=2a^2\\x^2+4=b^2\\3\sqrt{x^3+4x}=3ab\end{matrix}\right.\)
pt trên được viết lại thành
\(2a^2+b^2-3ab=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=\dfrac{1}{2}b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{x^2+4}\\\sqrt{x}=\dfrac{1}{2}\sqrt{x^2+4}\end{matrix}\right.\)
Đến đây dễ rồi nhé ^^
a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)
ĐKXĐ : \(x\inℝ\)
\(\Leftrightarrow\frac{x-1-\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}+\frac{x-1+\sqrt{x^2-2x+3}}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}=\frac{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}{\left(x-1+\sqrt{x^2-2x+3}\right)\left(x-1-\sqrt{x^2-2x+3}\right)}\)
\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)
\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)
\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)
\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)
\(\Leftrightarrow2x-2=-2\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)
Vậy phương trình có nghiệm duy nhất x = 0
\(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\\ \Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{52\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}-\dfrac{52\left(x+1\right)}{52\left(x+1\right)\left(x+10\right)}=\dfrac{9\left(x+1\right)\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow52\left(x+10\right)-52\left(x+1\right)=9\left(x+1\right)\left(x+10\right)\\ \Leftrightarrow9\left(x^2+10x+x+10\right)=52\left(x+10-x-1\right)\\ \Leftrightarrow9\left(x^2+11x+10\right)=52\cdot9\\ \Leftrightarrow x^2+11x+10=52\\ \Leftrightarrow x^2+14x-3x-42=0\\ \Leftrightarrow x\left(x+14\right)-3\left(x+14\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+14\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\left(T/m\right)\)
Vậy.............