GIẢI PT:
\(\left(\frac{x-1}{x}\right)^2+\left(\frac{x-1}{x+2}\right)^2=\frac{40}{9}\)
\(\frac{\left(x-1\right)^2}{x^2}+\frac{\left(x-1\right)^2}{\left(x-2\right)^2}=\frac{40}{49}\)
Giải PT
Giải pt
A,x2 + \(\frac{8+x^2}{\left(9+x\right)^2}\)= 40 b, \(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)
help me
#mã mã#
Giải PT : \(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\)
ĐKXĐ: x ≠ \(\pm\) 1
Từ phương trình ban đầu suy ra:
\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)
⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)
⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)
⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)
⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)
Vậy ...
giải hộ mk bài này nha????
giải phương trình :
1)\(5\left(\frac{x^2-4}{x^2-1}\right)-\left(\frac{x+2}{x-1}\right)^2-\left(\frac{x-2}{x+1}\right)^2=0\)
2)\(x^2+\left(\frac{x}{x-1}\right)^2=8\)
3)\(x^2+\left(\frac{81x^2}{\left(x+9\right)^2}\right)=40\)
4)\(\frac{\left(x-1\right)^2}{x^2}+\frac{\left(x-1\right)^2}{\left(x-2\right)^2}=\frac{40}{49}\)
5)\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)
giúp nha!!!!
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)2\)
giải pt
\(\left(x+4\right)^2\)nhấn lộn.mn giúp đỡ
giải pt
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Giải PT
\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\)
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha.
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2=\left(x+4\right)^2\)
giải pt trên
ĐKXĐ: \(x\ne0\)
Ta có \(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)
Đặt \(x^2+\frac{1}{x^2}=a\Rightarrow\left(x+\frac{1}{x}\right)^2=a+2\) pt trở thành:
\(8\left(a+2\right)+4a^2-4a\left(a+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8a+16+4a^2-4a^2-8a=\left(x+4\right)^2\)
\(\Leftrightarrow\left(x+4\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)
Giải hệ pt :
\(\left\{{}\begin{matrix}\left(x+y\right)\left(1+\frac{1}{xy}\right)=5\\\left(x^2+y^2\right)\left(1+\frac{1}{x^2y^2}\right)=9\end{matrix}\right.\)
ĐKXĐ: ...
\(\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=13\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^2+v^2=13\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\\left(u+v\right)^2-2uv=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\uv=6\end{matrix}\right.\)
Theo Viet đảo, u và v là nghiệm của: \(t^2-5t+6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+\frac{1}{x}=3\\y+\frac{1}{y}=2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow...\)