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\(\frac{x^2}{\left(x+2\right)}=3x^2-6x-3,x\ne-2\)
\(\Rightarrow x^2=\left(3x^2-6x-3\right)\left(x+2\right)^2\)
\(\Rightarrow x^2-\left(3x^2-6x-3\right)\left(x+2\right)^2=0\)
\(\Rightarrow x^2-\left(3x^4+12x^3+12x^2-6x^3-24x^2-24x-3x^2-12x-12\right)=0\)
\(\Rightarrow x^2-\left(3x^4+6x^3-15x^2-36x-12\right)=0\)
\(\Rightarrow16x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-2x^2+18x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-x^2\left(3x^2+6x+2\right)+\left(3x^2+6x+2\right)=0\)
\(\Rightarrow-\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(3x^2+6x=2\right)=0\\x^2-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{-3+\sqrt{3}}{3}\\\frac{-3-\sqrt{3}}{3},x\ne-2\\x=-\sqrt{6}\\x=\sqrt{6}\end{matrix}\right.\)
\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)
Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\)
\(Pt\Leftrightarrow10a^2+b^2-11ab=0\\ \Leftrightarrow10a^2-10ab-ab+b^2=0\\ \Leftrightarrow10a\left(a-b\right)-b\left(a-b\right)=0\\ \Leftrightarrow\left(10a-b\right)\left(a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}10a-b=0\\a-b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}10a=b\\a=b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{10\left(x-2\right)}{x+1}=\frac{x+2}{x-1}\left(1\right)\\\frac{x-2}{x+1}=\frac{x+2}{x-1}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow10\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow10\left(x^2-3x+2\right)=x^2+3x+2\\ \Leftrightarrow9x^2-33x+18=0\\ \Leftrightarrow9x^2-27x-6x+18=0\\ \Leftrightarrow9x\left(x-3\right)-6\left(x-3\right)=0\\ \Leftrightarrow\left(9x-6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\9x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{2}{3}\end{matrix}\right.\left(Tm\right)\)
\(\left(2\right)\Leftrightarrow\left(x-2\right)\left(x-1\right)=\left(x+1\right)\left(x+2\right)\\ \Leftrightarrow x^2-3x+2=x^2+3x+2=0\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\left(Tm\right)\)
Vậy pt có tập nghiệm \(S=\left\{0;3;\frac{2}{3}\right\}\)
\(\text{b) }\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}=12\left(\frac{x-2}{x-4}\right)^2\\ DKXD:x\ne2;x\ne4\\ \Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-2}\cdot\frac{x-2}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)
Đặt \(\frac{x+1}{x-2}=a;\frac{x-2}{x-4}=b\)
\(Pt\Leftrightarrow a^2+ab-12b^2=0\\ \Leftrightarrow a^2+4ab-3ab-12b^2=0\\ \Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\\ \Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3b\\a=-4b\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x+1}{x-2}=\frac{3\left(x-2\right)}{x-4}\left(1\right)\\\frac{x+1}{x-2}=\frac{-4\left(x-2\right)}{x-4}\left(2\right)\end{matrix}\right.\)
Tự giải tiếp nha.
a,Bạn xét 3 th
th1: x>=-1
th2: 1>x>-1
th3:x<=1
rồi trong từng th bạn bỏ dấu gttd và giải
b, \(\frac{x^2}{3}+\frac{48}{x^2}=10\left(\frac{x}{3}-\frac{4}{x}\right)\)
tương đương \(x^2+\frac{144}{x^2}=10\left(x-\frac{12}{x}\right)\)(nhân cả 2 vế với 3)
tương đương \(\left(x-\frac{12}{x}\right)^2+24-10\left(x-\frac{12}{x}\right)\)=0
đặt (x-12/x)=a
khi đó a^2-10a+24=0
giải a rồi tìm x thôi
c, đặt \(\sqrt[3]{x}\)=a
khi đó ta có 2a^2-5a=3
giải a rồi tìm x thôi
Chúc bạn học tốt!