giúp nhanh câu này vs mn ơi
K
Khách
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6 tháng 3 2022
\(P=\left(\dfrac{x^2+1}{x^2-9}-\dfrac{x}{x+3}+\dfrac{5}{3-x}\right):\left(\dfrac{2x+10}{x+3}-1\right)\)
\(=\left(\dfrac{x^2+1}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10}{x+3}-\dfrac{x+3}{x+3}\right)\)
\(=\left(\dfrac{x^2+1-x^2+3x-5x-15}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\left(\dfrac{-2x-14}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x+7}{x+3}\right)\)
\(=\dfrac{-2\left(x+7\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+7}\)
\(=\dfrac{-2}{x-3}\)
6 tháng 3 2022
đk : x khác -3 ; 3 ; -7
\(P=\left(\dfrac{x^2+1+x\left(x-3\right)+5x+15}{x^2-9}\right):\left(\dfrac{2x+10-x-3}{x+3}\right)\)
\(=\dfrac{2x^2+1+2x+15}{x^2-9}:\dfrac{x+7}{x+3}=\dfrac{2x^2+2x+16}{\left(x-3\right)\left(x+7\right)}\)
KC
2
NT
1
DT
21 tháng 2 2022
e tk:
https://vietjack.com/van-mau-lop-7/images/so-do-tu-duy-duc-tinh-gian-di-cua-bac-ho-a01.PNG
31 tháng 10 2021
a)\(2Ca+O_2\underrightarrow{t^o}2CaO\)
\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
\(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
29 tháng 10 2021
2: Để (d)//y=(m2+1)x-4 thì \(\left\{{}\begin{matrix}m^2=1\\m-5\ne-4\end{matrix}\right.\Leftrightarrow m=1\)
NT
1
10 tháng 3 2022
\(a)P=\left(\dfrac{x^2+2}{x^3-1}+\dfrac{x+1}{x^2+x+1}+\dfrac{1}{1-x}\right).\left(\dfrac{x^2}{x+1}+1\right).\left(x\ne1;x\ne-1\right).\\ P=\dfrac{x^2+2+x^2-1-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}.\dfrac{x^2+x+1}{x+1}.\\ P=\dfrac{x^2-x}{x-1}.\dfrac{1}{x+1}.\\ P=\dfrac{x\left(x-1\right)}{x-1}.\dfrac{1}{x+1}.\\ P=x.\dfrac{1}{x+1}.\\ P=\dfrac{x}{x+1}.\)
\(P=\dfrac{1}{4}.\Rightarrow\dfrac{x}{x+1}=\dfrac{1}{4}.\\ \Leftrightarrow4x-x-1=0.\\ \Leftrightarrow3x-1=0.\\ \Leftrightarrow x=\dfrac{1}{3}\left(TM\right).\)
LV
2
\(ĐK:x\ne0\)
\(\dfrac{x-1}{3}+\dfrac{x+3}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)+3\left(x+3\right)}{3x}=\dfrac{6x}{3x}\)
\(\Leftrightarrow x\left(x-1\right)+3\left(x+3\right)=6x\)
\(\Leftrightarrow x^2-x+3x+9-6x=0\)
\(\Leftrightarrow x^2-4x+9=0\)
Ta có: \(x^2-4x+9=x^2-4x+4+5=\left(x-2\right)^2+5\ge5>0\)
Vậy pt vô nghiệm