Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x^2}{\left(x+2\right)}=3x^2-6x-3,x\ne-2\)
\(\Rightarrow x^2=\left(3x^2-6x-3\right)\left(x+2\right)^2\)
\(\Rightarrow x^2-\left(3x^2-6x-3\right)\left(x+2\right)^2=0\)
\(\Rightarrow x^2-\left(3x^4+12x^3+12x^2-6x^3-24x^2-24x-3x^2-12x-12\right)=0\)
\(\Rightarrow x^2-\left(3x^4+6x^3-15x^2-36x-12\right)=0\)
\(\Rightarrow16x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-2x^2+18x^2-3x^4-6x^3+36x+12=0\)
\(\Rightarrow-x^2\left(3x^2+6x+2\right)+\left(3x^2+6x+2\right)=0\)
\(\Rightarrow-\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\left(3x^2+6x=2\right)=0\\x^2-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{-3+\sqrt{3}}{3}\\\frac{-3-\sqrt{3}}{3},x\ne-2\\x=-\sqrt{6}\\x=\sqrt{6}\end{matrix}\right.\)
\(\text{a) }10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{x^2-4}{x^2-1}=0\\ DKXD:x\ne-1;x\ne1\\ \Leftrightarrow10\left(\frac{x-2}{x+1}\right)^2+\left(\frac{x+2}{x-1}\right)^2-11\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-1\right)}=0\)
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)\(\frac{1}{a+b-x}\)=\(\frac{1}{a}\)+\(\frac{1}{b}\)-\(\frac{1}{x}\)\(\Leftrightarrow\)\(\frac{1}{a+b-x}\)+\(\frac{1}{x}\)=\(\frac{a+b}{ab}\)\(\Leftrightarrow\)\(\frac{x+a+b-x}{x\left(a+b-x\right)}\)=\(\frac{a+b}{ab}\)
\(\Leftrightarrow\)\(\frac{a+b}{xa+xb-x^2}\)=\(\frac{a+b}{ab}\)\(\Leftrightarrow\)\(xa+xb-x^2\)=\(ab\)\(\Leftrightarrow\)\(xa+xb-x^2-ab\)=\(0\)
\(\Leftrightarrow\)\(a\left(x-b\right)-x\left(x-b\right)=0\)\(\Leftrightarrow\)\(\left(x-b\right)\left(a-x\right)=0\)\(\Leftrightarrow\)\(x=b;x=a\)
b) \(\Leftrightarrow\)\(\frac{1}{\left(x+a-1\right)\left(x+a+1\right)}+\frac{1}{\left(x+a+1\right)\left(x-a+1\right)}\)=\(\frac{1}{\left(x-a-1\right)\left(x+a+1\right)}+\frac{1}{\left(x-a+1\right)\left(x+a-1\right)}\)\(\Leftrightarrow\)\(\frac{1}{\left(x+a-1\right)\left(x+a+1\right)}-\frac{1}{\left(x-a-1\right)\left(x+a+1\right)}\)=\(\frac{1}{\left(x-a+1\right)\left(x+a-1\right)}-\frac{1}{\left(x+a+1\right)\left(x-a+1\right)}\)\(\Leftrightarrow\)\(\frac{1}{\left(x+a+1\right)}\left(\frac{1}{x+a-1}-\frac{1}{x-a-1}\right)\)=\(\frac{1}{x-a+1}\left(\frac{1}{x+a-1}-\frac{1}{x+a+1}\right)\)\(\Leftrightarrow\)\(\frac{1}{x+a+1}.\frac{-2a}{\left(x+a-1\right)\left(x-a-1\right)}=\frac{1}{x-a+1}.\frac{2}{\left(x+a-1\right)\left(x+a+1\right)}\)(Quy dong phan so ttrong dau ngoac)
\(\Leftrightarrow\)\(\frac{-2a}{x-a-1}=\frac{2}{x-a+1}\)\(\Leftrightarrow\)\(-2a\left(x-a+1\right)=2\left(x-a-1\right)\)\(\Leftrightarrow\)\(-ax+a^2-a=x-a-1\)\(\Leftrightarrow\)\(-ax-x+a^2-1=0\)\(\Leftrightarrow\)\(\left(a+1\right)\left(-x+a-1\right)=0\)
neu a+1=0 thi phuong trinh co vo so nghiem, neu a+1\(\ne\)0 thi x=a-1