Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐK: x khác -1 và x khác 1.
\(PT\Leftrightarrow\frac{7x.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5x.\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x+21}{\left(x-1\right)\left(x+1\right)}=0\)
<=> 7x2 + 7x - 5x2 + 5x + x + 21 = 0
<=> 2x2 + 13x + 21 = 0
<=> 2x2 + 6x + 7x + 21 = 0
<=> 2x.(x + 3) + 7.(x + 3) = 0
<=> (x + 3).(2x + 7) = 0
<=> x + 3 = 0 hoặc 2x + 7 = 0
<=> x = -3 hoặc x = -7/2
Vậy S = {-7/2; -3}.
\(ĐKXĐ:x\ne2;x\ne4\)
\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)
\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)
\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)
\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)
\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)
\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)
Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)
a)\(\Leftrightarrow-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=-\frac{3x^2}{x+1}+\frac{3x}{x+1}+3x\)
\(\Rightarrow\frac{3x^2}{x+1}-\frac{4x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}-3x+\frac{1}{x-1}=0\)
\(\Leftrightarrow-\frac{2x\left(3x-5\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\int^{\frac{x-1}{1}=0}_{\frac{x+1}{1}=0}\Rightarrow x=0\)
=>3x=5
\(\Rightarrow x=\frac{3}{5}\)
vậy \(x=\frac{3}{5}\) hoặc 0
b)x = -(20309916*i+23555105)/9277755;
x = -(985155752*i-35635815)/916564140;
x = (985155752*i+35635815)/916564140;
x = (20309916*i-23555105)/9277755;
c)\(\Leftrightarrow\frac{x+2}{x-1}=\frac{1}{1}\Rightarrow\left(x+2\right)1=\left(x-1\right)1\)
vì \(\left(x+2\right)1\ne\left(x-1\right)1\)
=>x vô nghiệm hoặc đề sai
điều kiền x # 0
đặt \(t=x+\frac{1}{x};đk:t\ge2\)=>\(x^2+\frac{1}{x^2}=t^2-2\)
Ta được phương trình mới ẩn t : \(t^2-2t-5=0\)
tự giải phương trình nhé. lấy nghiệm t>= 2
\(\frac{x-3}{11}+\frac{x+1}{3}=\frac{x+7}{9}-1\)
\(\Leftrightarrow\frac{9\left(x-3\right)}{99}+\frac{33\left(x+1\right)}{99}=\frac{11\left(x+7\right)}{99}-\frac{99}{99}\)
\(\Leftrightarrow\frac{9\left(x-3\right)+33\left(x+1\right)}{99}=\frac{11\left(x+7\right)-99}{99}\)
\(\Leftrightarrow9\left(x-3\right)+33\left(x+1\right)=11\left(x+7\right)-99\)
\(\Leftrightarrow9x-27+33x+33=11x+77-99\)
\(\Leftrightarrow42x+6=11x-22\Leftrightarrow42x-11x=-6-22\)
\(\Leftrightarrow31x=-28\Leftrightarrow x=-\frac{28}{31}\)
Vậy phương trình có tập nghiệm S={-28/31}
\(\left(\frac{1}{x-1}+\frac{1}{x-4}\right)-\left(\frac{1}{x-2}+\frac{1}{x-3}\right)=0\)
\(\Leftrightarrow\frac{x-4+x-1}{\left(x-1\right).\left(x-4\right)}-\frac{x-3-x-2}{\left(x-2\right).\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x-5}{x^2-5x+4}-\frac{2x-5}{x^2-5x+6}=0\)
\(\Leftrightarrow\left(2x-5\right).\left(\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x^2-5x+4=x^2-5x+6\left(loai\right)\end{cases}}}\)
Vậy..