Giải PT:
\(x^2+\left(\frac{x}{x+1}\right)^2\)=1
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2020
\(8\left(x+\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8\left(x+\frac{1}{x}\right)^2-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)
\(\Leftrightarrow8\left(x^2+\frac{1}{x^2}+2\right)-8\left(x^2+\frac{1}{x^2}\right)=\left(x+2\right)^2\)
\(\Leftrightarrow\left(x+2\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 4 2019
ĐKXĐ: \(x\ne0\)
Ta có \(\left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2\)
Đặt \(x^2+\frac{1}{x^2}=a\Rightarrow\left(x+\frac{1}{x}\right)^2=a+2\) pt trở thành:
\(8\left(a+2\right)+4a^2-4a\left(a+2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow8a+16+4a^2-4a^2-8a=\left(x+4\right)^2\)
\(\Leftrightarrow\left(x+4\right)^2=16\)
\(\Rightarrow\left[{}\begin{matrix}x+4=4\\x+4=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)
30 tháng 3 2020
ĐK: \(x\in R\backslash\left\{-4,-3,-2,-1\right\}\)
PT ban đầu
\(\Leftrightarrow\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}+\frac{x+3-x-2}{\left(x+2\right)\left(x+3\right)}+\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}+\frac{x+5-x-4}{\left(x+4\right)\left(x+5\right)}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}=\frac{1}{x+1}-403\\ \Leftrightarrow\frac{1}{x+5}=403\\ \Leftrightarrow x+5=\frac{1}{403}\Leftrightarrow x=\frac{-2014}{403}\)
Chúc bạn học tốt nha
.
30 tháng 3 2020
Sr bạn nha, nhưng điều kiện là \(x\in R\backslash\left\{-5,-4,-3,-2,-1\right\}\). (Xét thiếu :>)
Chúc bạn học tốt nha.
OW
7
KN
11 tháng 2 2020
\(ĐKXĐ:x\ne-1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Rightarrow\frac{x-2+5x+5}{\left(x+1\right)\left(2-x\right)}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Rightarrow x-2+5x+5=15\)
\(\Rightarrow6x+3=15\Leftrightarrow6x=12\Leftrightarrow x=2\)
Vậy x = 2
NH
1
VH
7 tháng 7 2019
ĐKXĐ: x ≠ \(\pm\) 1
Từ phương trình ban đầu suy ra:
\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)
⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)
⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)
⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)
⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)
Vậy ...
2 tháng 2 2019
a,\(\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2=90\)\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+2.\frac{x}{x+1}.\frac{x}{x-1}+\left(\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}=90\)\(\Leftrightarrow\left(\frac{x^2-x+x^2+x}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}=90\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-90=0\)\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}-10\right)\left(\frac{2x^2}{x^2-1}+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2x^2}{x^2-1}=10\\\frac{2x^2}{x^2-1}=-9\end{cases}\Leftrightarrow......}\)
b,Đặt \(\frac{x-2}{x+1}=a;\frac{x+2}{x-1}=b\Rightarrow ab=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2-4}{x^2-1}\)
Từ đó ta có phương trình:\(20a^2-5b^2+48ab=0\Leftrightarrow20a^2-2ab-5b^2+50ab=0\)
\(\Leftrightarrow2a\left(10a-b\right)+5b\left(10a-b\right)=0\Leftrightarrow\left(2a+5b\right)\left(10a-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2a=-5b\\10a=b\end{cases}}\)
TH1:\(2a=-5b\Leftrightarrow\frac{2\left(x-2\right)}{x+1}=\frac{-5\left(x+2\right)}{x-1}\)\(\Rightarrow2\left(x-2\right)\left(x-1\right)=-5\left(x+2\right)\left(x+1\right)\)\(\Leftrightarrow2x^2-6x+4=-5x^2-15x-10\)\(\Leftrightarrow7x^2+9x+14=0\)
\(\Leftrightarrow7\left(x^2+\frac{9}{7}x+2\right)=0\Leftrightarrow7\left(x^2+2.\frac{9}{14}+\frac{81}{196}\right)+\frac{311}{28}=0\)
\(\Leftrightarrow7\left(x+\frac{9}{14}\right)^2+\frac{311}{28}=0\),vô lí
TH2:Tự làm nhé ,tương tự
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 2 2020
ĐKXĐ: ...
\(\Leftrightarrow\frac{2\left(x^2+1\right)}{\left(1-x^2\right)^2}+\frac{1}{4x^2}=\frac{\left(3x^2+1\right)^2}{144}\)
Đặt \(\left\{{}\begin{matrix}1-x^2=a\\4x^2=b\end{matrix}\right.\)
\(\Rightarrow\frac{2a+b}{a^2}+\frac{1}{b}=\frac{\left(a+b\right)^2}{144}\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{a^2b}=\frac{\left(a+b\right)^2}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\left(vn\right)\\a^2b=144\end{matrix}\right.\)
\(\Leftrightarrow\left(1-x^2\right)^2.4x^2=144\)
\(\Leftrightarrow\left(2x-2x^3\right)^2=12^2\)
\(\Leftrightarrow...\)
đk: \(x\ne-1\)
\(PT\Leftrightarrow x^2-\frac{2x^2}{x+1}+\left(\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\)
\(\Leftrightarrow\left(x-\frac{x}{x+1}\right)^2+2\frac{x^2}{x+1}=1\)
\(\Leftrightarrow\left(\frac{x^2}{x+1}+1\right)^2=2\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+\left(1-\sqrt{2}\right)x+\left(1-\sqrt{2}\right)=0\\x^2+\left(1+\sqrt{2}\right)x+\left(1+\sqrt{2}\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{\sqrt{2}-1+\sqrt{2\sqrt{2}-1}}{2}\\\frac{\sqrt{2}-1-\sqrt{2\sqrt{2}-1}}{2}\end{cases}\left(TM\right)}\)
Theo hđt : \(\left(a-b\right)^2=a^2-2ab+b^2\Rightarrow a^2+b^2=\left(a-b\right)^2+2ab\)
pt có dạng : \(\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\)ĐK : \(x\ne1\)
\(\Leftrightarrow\left(\frac{x^2+x-x}{x+1}\right)^2+\frac{2x^2}{x+1}=1\Leftrightarrow\frac{x^4+2x^2}{x+1}=\frac{x+1}{x+1}\)
\(\Rightarrow x^4+2x^2-x-1=0\Rightarrow x_1=-0,48...;x_2=0,82...\)( tmđk )