NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
18 tháng 6 2017
b) \(\frac{x-3}{x-2}+\frac{x+2}{x-4}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{x^2-7x+12+x^2-4}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Rightarrow\frac{2x^2-7x+8}{\left(x-2\right)\left(x-4\right)}=-1\)
.................
18 tháng 6 2017
a) \(\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(2x+3\right)\left(x-1\right)}{\left(x+1\right)\left(x^2+x+1\right)}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)}{x^3-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\)
\(\Rightarrow\left(x^3-1\right)\left[2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)\right]=\left(x^3-1\right)\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)=\left(2x-1\right)\left(2x+1\right)\)
\(\Rightarrow2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(2x-1\right)\left(2x+1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-\left(4x^2-1\right)=0\)
\(\Rightarrow2x^2+2x+2+2x^2-2x+3x-3-4x^2+1=0\)
\(\Rightarrow3x=0\)
\(\Rightarrow luon-dung-voi-moi-x\)
CH
Cô Hoàng Huyền
Admin
VIP
28 tháng 2 2018
a) \(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}=\frac{6}{x^2-2x+4}\)
Đặt \(x^2-2x+3=t\left(t\ge2\right)\), khi đó phương trình trở thành:
\(\frac{1}{t-1}+\frac{2}{t}=\frac{6}{t+1}\)
\(\Leftrightarrow\frac{t\left(t+1\right)+t^2-1}{\left(t-1\right)t\left(t+1\right)}=\frac{6t\left(t-1\right)}{\left(t-1\right)t\left(t+1\right)}\)
\(\Leftrightarrow t\left(t+1\right)+t^2-1=6t\left(t-1\right)\)
\(\Leftrightarrow2t^2+t-1=6t^2-6t\)
\(\Leftrightarrow-4t^2+7t-1=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=\frac{7+\sqrt{33}}{8}\\t=\frac{7-\sqrt{33}}{8}\end{cases}}\left(ktmđk\right)\)
Vậy phương trình vô nghiệm.
7 tháng 2 2020
\(ĐKXĐ:x\ne\pm2\)
\(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{2x-22}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-7x-2-2x+22=0\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
HN
0
S
1
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 4 2019
\(x\ne2;4\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)\left(x-2\right)=-\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x+12+x^2-4x+4+x^2-6x+8=0\)
\(\Leftrightarrow3x^2-17x+24=0\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{8}{3}\end{matrix}\right.\)
CW
3 tháng 2 2017
\(\frac{x}{x-1}=\frac{x+4}{x+1}\Leftrightarrow x^2+x-\left(x^2+3x-4\right)=0\)
\(\Leftrightarrow-2x+4=0\Leftrightarrow x=2\)
-----------------------------------------------------------------
\(\frac{3}{x-2}=\frac{2x-1}{x-2}-x\Leftrightarrow\frac{3}{x-2}=\frac{2x-1}{x-2}-\frac{x^2-2x}{x-2}\)
\(\Leftrightarrow2x-1-x^2+2x-3=0\Leftrightarrow-\left(x-2\right)^2=0\Leftrightarrow x=2\)
14 tháng 2 2020
\(ĐKXĐ:x\ne\pm1\)
\(pt\Leftrightarrow\frac{\left(x+1\right)\left(x^2+x+1\right)-3x^2\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)\(=\frac{2x\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)-3x^2\left(x^2+x+1\right)\)\(=2x\left(x+1\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x+1-3x^2\right)\left(x^2+x+1\right)\)\(=2x\left(x^2-1\right)\)
\(\Leftrightarrow-3x^4-2x^3-x^2+2x+1\)\(=2x^3-2x\)
\(\Leftrightarrow-3x^4-4x^3-x^2+4x+1=0\)
x - 3 / x -2 - x - 2 /x -4 =16/5
x - 3 / x - 2 - x - 2 /x -4 - 16/5 = 0
-16^2 +81x -88/ 5(x-2)(x-4) = 0
-16^2 +81x -81 =0
16^2 -81x +88 =0
x = -(-81) ± √(-81)^2 -4 *16 *88 /2*16
x = 81±√ 929/32
x1 =81+√929/32
x-2 =81-√929/32