Tìm x, biết:
a) \(70.\dfrac{4x+720}{x}=\dfrac{1}{2}\)
b) \(x^2+5x< 0\)
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LM
1
7 tháng 3 2022
a, đk : x khác -2 ; 2
\(\left(x+2\right)^2-8x=0\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(ktm)
pt vô nghiệm
b, đk : x khác -1 ; 1
\(x\left(x+1\right)-5x+3=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x=1\left(ktm\right);x=3\left(tm\right)\)
28 tháng 2 2021
`a,3x^2+7x+2=0`
`<=>3x^2+6x+x+2=0`
`<=>3x(x+2)+x+2=0`
`<=>(x+2)(3x+1)=0`
`<=>x=-2\or\x=-1/3`
28 tháng 2 2021
d) Ta có: (x-1)(x+2)=70
\(\Leftrightarrow x^2+2x-x-2-70=0\)
\(\Leftrightarrow x^2+x-72=0\)
\(\Leftrightarrow x^2+9x-8x-72=0\)
\(\Leftrightarrow x\left(x+9\right)-8\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)
Vậy: S={8;-9}
DH
6 tháng 7 2017
a, \(-\dfrac{1}{4}-\dfrac{3}{4}:x=-\dfrac{11}{36}\)
\(\Rightarrow\dfrac{3}{4}:x=-\dfrac{1}{4}-\left(-\dfrac{11}{36}\right)=\dfrac{1}{18}\)
\(\Rightarrow x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{27}{2}\)
b, \(70:\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{4x+720}{x}=140\)
\(\Rightarrow4x+720=140x\Rightarrow140x-4x=720\)
\(\Rightarrow136x=720\Rightarrow x=\dfrac{90}{17}\)
Chúc bạn học tốt!!!
QD
6 tháng 7 2017
a)\(\dfrac{-1}{4}-\dfrac{3}{4}:x=\dfrac{-11}{36}\)
\(\dfrac{3}{4}:x=\dfrac{-1}{4}-\left(\dfrac{-11}{36}\right)=\dfrac{1}{18}\)
\(\Rightarrow x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{27}{2}\)
b)\(70:\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\dfrac{4x+720}{x}=70:\dfrac{1}{2}=140\)
\(\Rightarrow4x+720=140x\)
\(\Rightarrow140x-4x=720\)
\(\Rightarrow136x=720\)
\(\Rightarrow x=\dfrac{90}{17}\)
6 tháng 4 2021
a) \(5x^4-4x^2-1=0\\ \Leftrightarrow5x^4+x^2-5x^2-1=0\\ \Leftrightarrow x^2\left(5x^2+1\right)-\left(5x^2+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(5x^2+1\right)=0\\ \Leftrightarrow x=\pm1\)
Vậy \(S=\left\{\pm1\right\}\) là nghiệm của pt
b) \(\dfrac{1}{x-4}-\dfrac{1}{x+4}=2\left(ĐKXĐ:x\ne\pm4\right)\\ \Leftrightarrow\dfrac{x+4-x+4}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x-4\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\\ \Rightarrow8=2x^2-32\\ \Leftrightarrow2x^2-40=0\\ \Leftrightarrow2\left(x-\sqrt{20}\right)\left(x+\sqrt{20}\right)=0\\ \Leftrightarrow x=\pm\sqrt{20}\left(tmđk\right)\)
Vậy \(S=\left\{\pm\sqrt{20}\right\}\) là nghiệm của pt
S
3
22 tháng 2 2022
a: \(\Leftrightarrow2x\left(x^2+2x+5\right)=0\)
=>x=0
b: \(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{x+1}{x-3}=\dfrac{1}{2}\)
\(\Leftrightarrow x^2-4x+3=2x\left(x-3\right)-2\left(x^2-1\right)\)
\(\Leftrightarrow x^2-4x+3=2x^2-6x-2x^2+2=-6x+2\)
\(\Leftrightarrow x^2+2x+1=0\)
=>x=-1(nhận)
22 tháng 2 2022
\(a,2x^3+4x^2+10x=0\\ \Leftrightarrow2x\left(x^2+2x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x^2+2x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x^2+2x+1\right)+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2+4=0\left(vô..lí\right)\end{matrix}\right.\)
\(b,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne3\\x\ne4\end{matrix}\right.\\ \dfrac{x^2-4x}{x^2-5x+4}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}-\dfrac{1}{2}=\dfrac{x+1}{x-3}\\ \Leftrightarrow\dfrac{x}{x-1}-\dfrac{1}{2}-\dfrac{x+1}{x-3}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2\left(x+1\right)\left(x-1\right)}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)}-\dfrac{x^2-4x+3}{2\left(x-1\right)\left(x-3\right)}-\dfrac{2x^2-2}{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-6x-x^2+4x-3-2x^2+2}{2\left(x-1\right)\left(x-3\right)}=0\)
\(\Rightarrow-x^2-2x-1=0\)
\(\Leftrightarrow x^2+2x+1=0\\ \Leftrightarrow\left(x+1\right)^2=0\\ \Leftrightarrow x+1=0\\ \Leftrightarrow x=-1\left(tm\right)\)
29 tháng 7 2021
a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)
\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)
hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)
b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)
nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)
hay \(x=\dfrac{8}{41}\)
c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)
\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)
\(\Leftrightarrow\left|2x-1\right|=11\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)
Giải:
a) \(70.\dfrac{4x+720}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{280x+50400}{x}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(280x+50400\right)=x\)
\(\Leftrightarrow560x+100800=x\)
\(\Leftrightarrow560x-x=-100800\)
\(\Leftrightarrow549x=-100800\)
\(\Leftrightarrow x=-\dfrac{11200}{61}\)
Vậy ...
b) \(x^2+5x< 0\)
\(\Leftrightarrow x\left(x+5\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< -5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0>x>-5\\x\in\varnothing\end{matrix}\right.\)
Vậy ...