Giải PT :\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
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BL
0
DF
1
HP
23 tháng 1 2021
ĐK: \(x\ne-5\)
\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
\(\Leftrightarrow x^2+\dfrac{25x^2}{\left(x+5\right)^2}-\dfrac{10x^2}{x+5}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\left(x-\dfrac{5x}{x+5}\right)^2+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow\dfrac{x^4}{\left(x+5\right)^2}+\dfrac{10x^2}{x+5}=11\)
\(\Leftrightarrow y^2+10y-11=0\left(y=\dfrac{x^2}{x+5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-11\end{matrix}\right.\)
TH1: \(y=1\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=1\)
\(\Leftrightarrow x^2=x+5\)
\(\Leftrightarrow x=\dfrac{1\pm\sqrt{21}}{2}\left(tm\right)\)
TH2: \(y=-11\)
\(\Leftrightarrow\dfrac{x^2}{x+5}=-11\)
\(\Leftrightarrow x^2=-11x-55\)
\(\Rightarrow\) vô nghiệm
Vậy phương trình đã cho có nghiệm \(x=\dfrac{1\pm\sqrt{21}}{2}\)
20 tháng 9 2021
1) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
2) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
LT
2
7 tháng 2 2021
ĐKXĐ : \(x\ne\pm2\)
PT \(\Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{x^2-4}=\dfrac{2\left(x-11\right)}{x^2-4}\)
\(\Leftrightarrow\left(x-2\right)^2-3\left(x+2\right)=2\left(x-11\right)\)
\(\Leftrightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\) ( TM )
Vậy ...
7 tháng 2 2021
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x-2}{2+x}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-7x-2-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={4;5}
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 2 2021
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
DT
11 tháng 4 2022
1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
Y
11 tháng 4 2022
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)Pt có 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)
Vậy \(S=\left\{4;-1\right\}\)
\(3,x^4-5x^2+6=0\)
Đặt \(t=x^2\left(t\ge0\right)\)
Pt trở thành
\(t^2-5t+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)
\(\Rightarrow\)Pt ó 2 nghiệm pb
\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)
\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)
Vậy \(S=\left\{\pm\sqrt{3}\right\}\)
14 tháng 2 2023
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)
\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)
=>\(-9x^4+81x^2-36=0\)
=>9x^4-81x^2+36=0
=>x^4-9x^2+4=0
=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)
=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)
\(\dfrac{x^2\left(x+5\right)^2+25x^2-11\left(x+5\right)^2}{\left(x+5\right)^2}=\dfrac{x^4+10x^3+50x^2-11x^2-11.2.5x-11.5.5}{\left(x+5\right)^2}\)
\(\dfrac{\left[x^4-x^3-5x^2\right]+\left[11x^3-11x^2-11.5x\right]+55x^2-55x-5.55}{\left(x+5\right)^2}\)
\(\dfrac{x^2\left(x^2-x-5\right)+11x\left(x^2-x-5\right)+55\left(x^2-x-5\right)}{\left(x+5\right)^2}=\dfrac{\left(x^2-x-5\right)\left(x^2+11x+55\right)}{\left(x+5\right)^2}\)
\(\left[{}\begin{matrix}x^2-x-5=0\left(1\right)\\x^2+11x+55=0\left(2\right)\end{matrix}\right.\)
(2) vô nghiệm
(1)\(\Leftrightarrow\) \(\left[x^2-\dfrac{1}{2}x+\dfrac{1}{4}\right]=\dfrac{21}{4}\Rightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{21}{4}\)
\(\left[{}\begin{matrix}x=\dfrac{1-\sqrt{21}}{2}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\)
\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\) (ĐKXĐ: \(x\ne-5\))
\(\Leftrightarrow\dfrac{x^2\left(x+5\right)^2+25x^2}{\left(x+5\right)^2}=\dfrac{11\left(x+5\right)^2}{\left(x+5\right)^2}\)
\(\Rightarrow x^4+10x^3+25x^2+25x^2-11x^2-110x-275=0\\ \Leftrightarrow x^4+10x^3+39x^2-110x-275=0\)
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