Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2:
b: \(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)=\left(x-y\right)\left(x+y+1\right)\)
\(\dfrac{1}{-2x^2+4x-2}=\dfrac{x-2}{-2\left(x-1\right)^2\left(x-2\right)}\\ \dfrac{1}{2x^2-6x+4}=\dfrac{x-1}{2\left(x-1\right)^2\left(x-2\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}m\ge n+4\\n+6\ge m\end{matrix}\right.\Rightarrow n+6\ge m\ge n+4\Rightarrow n+5=m\\ \Rightarrow2\left(m-n\right)+3=2\left(n+5-n\right)+3=13\)
\(ĐK:x\ne0;2\)
\(\Rightarrow\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) ĐKXĐ: x≠0, x≠2
<=>\(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
=>x(x+2) - x + 2 =2
<=>x\(^2\) + 2x - x = 2-2
<=> x\(^2\) + x =0
<=> x(x + 1) = 0
<=>\(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
mà ĐKXĐ: x≠0, x≠2
Vậy x = -1
a: Xét tứ giác ABCM có
MC//AB
MC=AB
Do đó: ABCM là hình bình hành
2.
Đk: \(x>0,x\ne1\)
\(B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x-1}\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
Vậy B=...
3.\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\) , \(\dfrac{A}{B}=\dfrac{x}{\sqrt{x}-1}\)
\(P< 0\Leftrightarrow\dfrac{x}{\sqrt{x}-1}< 0\) \(\Leftrightarrow\sqrt{x}-1< 0\left(dox>0\right)\)
\(\sqrt{x}< 1\Leftrightarrow0< x< 1\)
Vậy \(0< x< 1\) thì P âm.
\(\dfrac{1}{-3x^2+18x-27}=\dfrac{-1}{3\left(x-3\right)^2}=\dfrac{-\left(x+1\right)}{3\left(x-3\right)^2\left(x+1\right)}\\ \dfrac{1}{3x^2-6x-9}=\dfrac{1}{3\left(x-3\right)\left(x+1\right)}=\dfrac{x-3}{3\left(x-3\right)^2\left(x+1\right)}\)
đk : x khác 1 ; -3
\(\Rightarrow2x+6+4x-4=3x+11\Leftrightarrow3x=9\Leftrightarrow x=3\left(tm\right)\)
