Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)^2+x^2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2\left[\left(x+1\right)^2-\left(x-1\right)^2\right]}{\left[\left(x-1\right)\left(x+1\right)\right]^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x^2-1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2.2.2x}{x^4-2x^2+1}=\frac{10}{9}\)
\(\Leftrightarrow36x^3=10x^4-20x^2+10\Leftrightarrow18x^3=5x^4-10x^2+5\Leftrightarrow5x^4-18x^3-10x^2\)+5=0
đến đây tự giải tiếp
ĐK:\(x\ne1;x\ne-1\)
\(pt\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2}{9\left(x-1\right)^2\left(x+1\right)^2}=0\)
\(\Leftrightarrow9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^4+18x^3+9x^2+9x^4-18x^3+9x^2-10x^4+20x^2-10=0\)
\(\Leftrightarrow8x^4+38x^2-10=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=5\left(l\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
ĐK: xy\(\ne\)0
HPT đã cho tương đương: \(\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=9\end{cases}}\)
Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=S\\\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)=P\end{cases}}\)
Hệ trở thành:
\(\hept{\begin{cases}S^2-2P=9\\S=5\end{cases}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{x}=2;y+\frac{1}{y}=3\\x+\frac{1}{x}=3;y+\frac{1}{y}=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1;y=\frac{3\pm\sqrt{5}}{2}\\x=\frac{3\pm\sqrt{5}}{2};y=1\end{cases}}}\)
Vậy HPT đã cho có nghiệm (x;y)=\(\left(1;\frac{3\pm\sqrt{5}}{2}\right);\left(\frac{3\pm\sqrt{5}}{2};1\right)\)
\(\hept{\begin{cases}\left(x+y\right)\left(1+\frac{1}{xy}\right)=5\\\left(x^2+y^2\right)\left(1+\frac{1}{x^2y^2}\right)=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+y+\frac{1}{x}+\frac{1}{x}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=9\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=13\end{cases}}\)
\(\left(x+\frac{1}{x};y+\frac{1}{y}\right)\rightarrow\left(a;b\right)\)
Hệ pt \(\Leftrightarrow\hept{\begin{cases}a+b=5\\a^2+b^2=13\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=5\\\left(a+b\right)^2-2ab=13\end{cases}\Leftrightarrow\hept{\begin{cases}a+b=5\\ab=6\end{cases}}}\)
Tự làm nốt nhé
bạn tham khảo thêm cách này nha Shonogeki No Soma
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\\x\ne-1\end{cases}}\)
Đặt \(a=\left(x-1\right)^3;b=x^3;c=\left(x+1\right)^3\)
pt đã cho đc viết lại thành
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}a=-b\\b=-c\\c=-a\end{cases}}\) (kí hiệu [..] mới đúng nha)
- TH1: a = -b hay \(\left(x-1\right)^3=-x^3\) \(\Leftrightarrow2x^3-3x^2+3x-1=0\) \(\Leftrightarrow x=\frac{1}{2}\) (Nhận)
- TH2: b = -c hay \(\left(x+1\right)^3=-x^3\) \(\Leftrightarrow2x^3+3x^2+3x+1=0\) \(\Leftrightarrow x=-\frac{1}{2}\) (Nhận)
- TH3: c = -a hay \(\left(x+1\right)^3=-\left(x-1\right)^3\) \(\Leftrightarrow x=0\) (Loại)
KL: \(S=\left\{\frac{1}{2};-\frac{1}{2}\right\}\)
\(\frac{1}{\left(x-1\right)^3}+\frac{1}{\left(x+1\right)^3}+\frac{1}{x^3}=\frac{1}{3x\left(x^2+2\right)}\)
\(\Leftrightarrow4x^8+15x^6+12x^4+8x^2-6=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x^2+3\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)
2) đặt \(x^2+x+1=t\left(t>0\right)\) ==> \(x^2+x+2=t+1\)
nên pt trên trở thành
\(\left(\frac{1}{t}\right)^2+\left(\frac{1}{t+1}\right)^2=\frac{13}{36}\)
<=> \(\frac{1}{t^2}+\frac{1}{t^2+2t+1}=\frac{13}{36}\)
<=> \(13t^4+26t^3-59t^2-72t-36=0\)
<=> \(13t^4-26t^3+52t^3-104t^2+45t^2-90t+18t-36=0\)
<=> \(13t^3\left(t-2\right)+52t^2\left(t-2\right)+45t\left(t-2\right)+18\left(t-2\right)=0\)
<=>\(\left(t-2\right)\left(13t^3+52t^2+45t+18\right)=0\)
<=> \(\left(t-2\right)\left(t+3\right)\left(13t^2+13t+6\right)=0\)
<=> \(\orbr{\begin{cases}t=2\left(tmdk\right)\\t=-3\left(ktmdk\right)\end{cases}}\)
đến đây bạn thay vào làm nốt nhá
1.
Đặt \(a=\frac{x\left(5-x\right)}{x+1};b=x+\frac{5-x}{x+1}\)
Ta cần giải pt : \(a.b=6\)(1)
Ta có: \(a+b=\frac{x\left(5-x\right)}{x+1}+x+\frac{5-x}{x+1}=\frac{5x-x^2+x^2+x+5-x}{x+1}=5\)
\(\Rightarrow a=5-b\)
Thế \(a=5-b\)vào (1)
\(\Rightarrow\left(5-b\right)b=6\)
\(\Leftrightarrow b^2-5b+6=0\)
\(\Leftrightarrow\left(b-2\right)\left(b-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}b=2\\b=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x+\frac{5-x}{x+1}=2\\x+\frac{5-x}{x+1}=3\end{cases}}}\)
Giải 2 pt trên, ta có nghiệm : \(x=1\)
Giải phương trình: \(\frac{1}{\left(x^2+x+1\right)^2}+\frac{1}{\left(x^2+x+2\right)^2}=\frac{5}{4}\)
Đặt \(x^2+x+1=a\)
\(pt\Leftrightarrow\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{5}{4}.\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{a+1}\right)^2+\frac{2}{a\left(a+1\right)}-\frac{5}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a\left(a+1\right)}\right)^2+\frac{2}{a\left(a+1\right)}-\frac{5}{4}=0\)
đặt \(\frac{1}{a\left(a+1\right)}=b\)
\(\Leftrightarrow b^2+2b-\frac{5}{4}=0\Leftrightarrow4b^2+8b-5=0\)
\(\left(2b-1\right)\left(2b+5\right)=0.\)
đến đây tự full đi.