Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ: \(x\ne0;\pm1\)
\(\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)^2}-\dfrac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\dfrac{1}{x}=2\)
\(\Leftrightarrow\dfrac{x^2+3x+2}{\left(x-1\right)^2}-\dfrac{x-2}{x-1}=2x\)
\(\Leftrightarrow\dfrac{x^2-2x+1+5x-5+6}{\left(x-1\right)^2}-\dfrac{x-1-1}{x-1}=2x\)
\(\Leftrightarrow1+\dfrac{5}{x-1}+\dfrac{6}{\left(x-1\right)^2}-1+\dfrac{1}{x-1}=2x\)
\(\Leftrightarrow\dfrac{3}{x-1}+\dfrac{3}{\left(x-1\right)^2}=x=x-1+1\)
Đặt \(\dfrac{1}{x-1}=a\) phương trình trở thành:
\(3a+3a^2=\dfrac{1}{a}+1=\dfrac{a+1}{a}\)
\(\Leftrightarrow3a\left(a+1\right)-\dfrac{a+1}{a}=0\)
\(\Leftrightarrow\left(a+1\right)\left(3a-\dfrac{1}{a}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\\3a=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=-1\\a=\dfrac{\sqrt{3}}{3}\\a=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{x-1}=-1\\\dfrac{1}{x-1}=\dfrac{\sqrt{3}}{3}\\\dfrac{1}{x-1}=\dfrac{-\sqrt{3}}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=1+\sqrt{3}\\x=1-\sqrt{3}\end{matrix}\right.\)
a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)
\(\Rightarrow\) vô nghiệm
b) ĐK: \(x\ne1;-1\)
\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)
\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)
a) Ta có: \(\Delta'=100-104=-4< 0\)
Vậy phương trình vô nghiệm.
b) ĐKXĐ: \(x\ne1;x\ne-1\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x-19=45x+45\)
\(\Leftrightarrow43x=-64\)
\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)
Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)
\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)
\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)
\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)
Bài 1:
Đặt: \(\left\{{}\begin{matrix}u=\dfrac{1}{2x-2}\\v=\dfrac{1}{y-1}\end{matrix}\right.\) (ĐK: \(x,y\ne1\))
Hệ trở thành:
\(\Leftrightarrow\left\{{}\begin{matrix}u-v=2\\3u-2v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3u-3v=6\\3u-2v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-v=5\\u-v=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=-5\\u=2+-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=-5\\u=-3\end{matrix}\right.\)
Trả lại ẩn của hệ pt:
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y-1}=-5\\\dfrac{1}{2x-2}=-3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=-\dfrac{1}{5}\\2x-2=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x=\dfrac{5}{6}\end{matrix}\right.\left(tm\right)\)
Đặt x/x+1=a; y/y+1=b
Hệ sẽ là 2a+b=căn 2 và a+3b=-1
=>2a+b=căn 2 và 2a+6b=-2
=>-5b=căn 2+2 và a=-1-3b
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{-\sqrt{2}-2}{5}\\a=-1-3\cdot\dfrac{-\sqrt{2}-2}{3}=-1+\sqrt{2}+2=1+\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y}{y+1}=\dfrac{-2-\sqrt{2}}{5}\\\dfrac{x}{x+1}=1+\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+1-1}{y+1}=\dfrac{-2-\sqrt{2}}{5}\\\dfrac{x+1-1}{x+1}=1+\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y+1}=1-\dfrac{-2-\sqrt{2}}{5}=1+\dfrac{2+\sqrt{2}}{5}=\dfrac{7+\sqrt{2}}{5}\\\dfrac{1}{x+1}=1-1-\sqrt{2}=-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{5}{7+\sqrt{2}}-1=\dfrac{5-7-\sqrt{2}}{7+\sqrt{2}}=\dfrac{-2-\sqrt{2}}{7+\sqrt{2}}\\x=-\dfrac{1}{\sqrt{2}}-1\end{matrix}\right.\)
\(x^2+2x\sqrt{x+\dfrac{1}{x}}=8x-1\left(x\ne0\right)\)
Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge\dfrac{1}{8}\)
Vì \(x\ne0\Rightarrow\) chia 2 vế cho x,ta được:
\(x+2\sqrt{x+\dfrac{1}{x}}=8-\dfrac{1}{x}\Rightarrow x+\dfrac{1}{x}+2\sqrt{x+\dfrac{1}{x}}=8\)
Đặt \(\sqrt{x+\dfrac{1}{x}}=a\left(a>0\right)\)
pt trở thành \(a^2+2a-8=0\Rightarrow a^2-2a+4a-8=0\)
\(\Rightarrow a\left(a-2\right)+4\left(a-2\right)=0\Rightarrow\left(a-2\right)\left(a+4\right)=0\)
mà \(a>0\Rightarrow a=2\Rightarrow\sqrt{x+\dfrac{1}{x}}=2\Rightarrow x+\dfrac{1}{x}=4\)
\(\Rightarrow\dfrac{x^2-4x+1}{x}=0\Rightarrow x^2-4x+1=0\)
\(\Delta=\left(-4\right)^2-4=12\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-\sqrt{12}}{2}=2-\sqrt{3}\\x-\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+\sqrt{12}}{2}=2+\sqrt{3}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{2-\sqrt{3};2+\sqrt{3}\right\}\)
\(DKXD:x\ne\pm1\\ pt\Rightarrow2x^2+x+1+2x-2=x^2-1\\ \Leftrightarrow x^2+3x=0\\ \Leftrightarrow x\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\left(N\right)\\ \Rightarrow S=\left\{0;-3\right\}\)