giải pt : \(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
Giải PT :\(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
\(\dfrac{x^2\left(x+5\right)^2+25x^2-11\left(x+5\right)^2}{\left(x+5\right)^2}=\dfrac{x^4+10x^3+50x^2-11x^2-11.2.5x-11.5.5}{\left(x+5\right)^2}\)
\(\dfrac{\left[x^4-x^3-5x^2\right]+\left[11x^3-11x^2-11.5x\right]+55x^2-55x-5.55}{\left(x+5\right)^2}\)
\(\dfrac{x^2\left(x^2-x-5\right)+11x\left(x^2-x-5\right)+55\left(x^2-x-5\right)}{\left(x+5\right)^2}=\dfrac{\left(x^2-x-5\right)\left(x^2+11x+55\right)}{\left(x+5\right)^2}\)
\(\left[{}\begin{matrix}x^2-x-5=0\left(1\right)\\x^2+11x+55=0\left(2\right)\end{matrix}\right.\)
(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: \(x\ne-5\))
\(\Leftrightarrow\dfrac{x^2\left(x+5\right)^2+25x^2}{\left(x+5\right)^2}=\dfrac{11\left(x+5\right)^2}{\left(x+5\right)^2}\)
\(\Rightarrow x^4+10x^3+25x^2+25x^2-11x^2-110x-275=0\\ \Leftrightarrow x^4+10x^3+39x^2-110x-275=0\)
mình ko biết phân tích sao nữa
Ta có: \(x^{2^{ }}+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
ĐKXĐ: x\(\ne\)-5
<=> \(x^2-2x\dfrac{5x}{x+5}+\dfrac{25x^2}{\left(x+5\right)^2}+2x\dfrac{5x}{x+5}=11\)
<=> \(\left(x-\dfrac{5x}{x+5}\right)^2+10\dfrac{x^2}{x+5}=11\)
<=> \(\left(\dfrac{x^2}{x+5}\right)^2+10\dfrac{x^2}{x+5}=11\)
Gọi \(\dfrac{x^2}{x+5}=t\)
=> t2+10t = 11
<=> t2+10t-11=0
<=> (t-1)(t+1)+10(t-1)=0
<=>(t-1)(t+11)=0
=>\(\left[{}\begin{matrix}t-1=0\\t+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-11\end{matrix}\right.\)
hay: \(\left[{}\begin{matrix}\dfrac{x^2}{x+5}=1\\\dfrac{x^2}{x+5}=-11\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{21}}{2}\\x=\dfrac{1+\sqrt{21}}{2}\end{matrix}\right.\)
giải phương trình \(x^2+\dfrac{25x^2}{\left(x+5\right)^2}=11\)
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 \(x=\dfrac{1\pm\sqrt{21}}{2}\)
help me!!!
Rút gọn: \(\sqrt{\left(\sqrt{3}+1\right)2^{ }}-\sqrt{4-2\sqrt{3}}\)
Giải PT: \(\dfrac{3}{5}\). \(\sqrt{25x-50}\) - \(\sqrt{x-2}\) = 6
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) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
giải phương trình
1)\(\dfrac{1}{x^2}+\dfrac{1}{\left(x+2\right)^2}=\dfrac{10}{9}\)
2) \(x^2+\dfrac{25x^2}{\left(x+5\right)^5}=11\)
3) x\(\left(\dfrac{5-x}{x+1}\right)\left(x+\dfrac{5-x}{x+1}\right)=6\)
4) \(\left(\dfrac{x}{x+1}\right)^2\left(\dfrac{x}{x-1}\right)^2=90\)
giải pt
\(\dfrac{x-2}{2+x}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
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 ...
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}
Giải pt sau: \(10\left(x+\dfrac{1}{x}\right)^2+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
(1) giải pt quy về \(ax^2+bx+c=0\)
1) \(x^2=3x\) 2) \(x^2-3x=4\)
3) \(x^4-5x^2+6=0\) 4) \(x^3=9x\)
5) \(\left(x+2\right)\left(x-3\right)=x^2-4\) 6) \(\dfrac{x+11}{x^2-1}-\dfrac{x-1}{x+1}=\dfrac{2\left(x+7\right)}{x+1}\)
giúp mk vs mk cần gấp
1)
<=> \(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
x= 0
x = 3
2) <=> \(x\left(x-3\right)=4\)
=> \(x=\dfrac{4}{x}+3\)
\(2,x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)
\(\Rightarrow\)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 \(S=\left\{\pm\sqrt{3}\right\}\)
\(4,x^3=9x\)
\(\Leftrightarrow x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
Vậy \(S=\left\{0;\pm3\right\}\)
\(5,\left(x+2\right)\left(x-3\right)=x^2-4\)
\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)
\(\Leftrightarrow-x-2=0\)
\(\Leftrightarrow-x=2\)
\(\Leftrightarrow x=-2\)
Vậy \(S=\left\{-2\right\}\)
Giải PT này giùm mk nha : \(10\cdot\left(\dfrac{x-2}{x+1}\right)^2+\left(\dfrac{x+2}{x-1}\right)^2-11\cdot\left(\dfrac{x^2-4}{x^2-1}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2}{\left(x^2-1\right)^2}-\dfrac{11\left(x^4-5x^2+4\right)}{\left(x^2-1\right)^2}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)^2+\left(x^2+3x+2\right)^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)^2-6x\left(x^2+2\right)+9x^2+\left(x^2+2\right)^2+6x\left(x^2+2\right)+9x^2-11\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow2\left(x^2+2\right)^2+18x^2-11x^4+55x^2-44=0\)
\(\Leftrightarrow2\left(x^4+4x^2+4\right)-11x^4+73x^2-44=0\)
=>\(-9x^4+81x^2-36=0\)
=>9x^4-81x^2+36=0
=>x^4-9x^2+4=0
=>\(x^2=\dfrac{9\pm\sqrt{65}}{2}\)
=>\(x=\pm\sqrt{\dfrac{9\pm\sqrt{65}}{2}}\)
Phân tích đa thức \(\dfrac{2}{5}x^2+5x^3+x^2y\) thành nhân tử
a. \(x^2\left(\dfrac{2}{5}+5x+y\right)\)
b. \(\dfrac{1}{5}x^2\left(2+25x+5y\right)\)
Cách phân tích nào đúng a hay b và GIẢI THÍCH VÌ SAO?
A. Cách B sai vì 5 : 2/5 thì ko thể nào = 25 đc.