NV
1
K
Khách
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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 2 2021
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)
\(\Leftrightarrow\left(x-5\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)
TP
2
D
16 tháng 1 2019
\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-x+2\right)\left(x-1+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right).1.\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)
\(TH1:x+1=0\)
\(\Leftrightarrow x=-1\)
\(TH2:2x-3=0\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy phương trình có tập nghiệm là:\(S=\left\{-1;\frac{3}{2}\right\}\)
16 tháng 1 2019
\(\left(x+1\right)\left(x-1\right)^2-\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1\right)^2-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x-1-x+2\right)\left(x-1+x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left[1.\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\2x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}\)
Vậy................
DT
0
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 4 2021
ĐKXĐ: ...
\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)
\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
K
giải pt sau \(\left(\dfrac{x+1}{x-2}\right)^2-3\left(\dfrac{2x-4}{x-4}\right)^2+\dfrac{x+1}{x-4}=0\)
1
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
ĐKXĐ: \(x\ne\left\{2;4\right\}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)
Phương trình trở thành:
\(a^2-12b^2+ab=0\)
\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)
\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)
Bạn tự quy đồng và hoàn thành phần còn lại nhé
HH
28 tháng 6 2018
Giải:
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
ĐKXĐ: \(x\ne\left\{1;2;3;4\right\}\)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-3\right)\left(x-2\right)}=\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-1\right)\left(x-4\right)}\)
\(\Rightarrow\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x-4\right)=\left(x-1\right)\left(x-2\right)+\left(x-2\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x-3\right)+\left(x-1\right)\right]=\left(x-2\right)\left[\left(x-1\right)+\left(x-3\right)\right]\)
\(\Leftrightarrow x-4=x-2\)
\(\Leftrightarrow0x=2\)
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
K
2
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2022
\(\Leftrightarrow1+3x+3x^2+x^3+1-3x+3x^2-x^3=6\left(x+1\right)^2\)
\(\Leftrightarrow6x^2+2=6\left(x^2+2x+1\right)\)
\(\Leftrightarrow6x^2+2=6x^2+12x+6\)
\(\Leftrightarrow12x=-4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
4T
10 tháng 1 2022
1+3x+3x2+1+1-3x2+3x-1=6(x2+2x+1)
6x2 +2 = 6x2+12x+6
-12x - 4 = 0
=> x = 3
Vì \(x^2+x+1=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
Vậy ....