Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(pt\Leftrightarrow-\frac{x}{2x^2-5}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=\frac{x}{2x^2+10x}-\frac{5}{2x^2+10x}\)
=>\(-\frac{x}{2x^2+10x}+\frac{5}{2x^2+10x}-\frac{x}{2x^2-50}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=0\)
\(\Leftrightarrow-\frac{5\left(x^2+8x-5\right)}{2\left(x-5\right)x\left(x^2-5\right)}=0\)
\(\Rightarrow\frac{1}{x-5}=0\Leftrightarrow\frac{1}{x}=0\Rightarrow\frac{1}{x^2-5}=0\)
=>x2+8x-5=0
=>82-(-4(1.5))=84
=>x1=(-8)+8:2=\(\sqrt{21}-4\)
=>x2=(-8)+8:2=\(-\sqrt{21}-4\)
=>x=±\(\sqrt{21}-4\)
b)\(\Leftrightarrow-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=\frac{16}{x^2-1}\)
\(\Rightarrow-\frac{16}{x^2-1}-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=0\)
\(\Rightarrow\frac{4\left(x-4\right)}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow\frac{1}{x-1}=0\Rightarrow\frac{1}{x+1}=0\)
=>x=4
c)\(\Leftrightarrow-\frac{x^2}{x+1}-\frac{x}{x+1}+\frac{2}{x+1}+x+2=\frac{x}{x+1}-\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}\)
\(\Rightarrow-\frac{x^2}{x+1}-\frac{2x}{x+1}+\frac{3}{x+1}-\frac{x}{x-1}+x-\frac{1}{x-1}+2=0\)
\(\Rightarrow\frac{2\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow\frac{1}{x-1}=0\Rightarrow\frac{1}{x+1}=0\)
=>x=3
c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)
\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)
\(\Rightarrow42x-6=60+18-10x\)
\(\Leftrightarrow52x-84=0\)
\(\Leftrightarrow x=\dfrac{21}{13}\)
Vậy....
d) tương tự
a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
Vậy....
\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}\right)=\left(x-23\right)\left(\frac{1}{26}+\frac{1}{27}\right)\text{ nhận thấy:}\frac{1}{24}+\frac{1}{25}>\frac{1}{26}+\frac{1}{27}\)
\(\Rightarrow x-23=0\Leftrightarrow x=23\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\Rightarrow\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)=\left(\frac{x+3}{2002}+1\right)+\left(\frac{x+4}{2001}+1\right)\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\text{dạng giống câu a rồi nha}\)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\Leftrightarrow300-x=0\)
Vậy: x=300
a) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
\(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)
\(\Leftrightarrow\)\(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)
\(\Leftrightarrow\)\(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-50=0\) (vì 1/50 + 1/49 + 1/48 + 1/47 + 1/25 > 0 )
\(\Leftrightarrow\)\(x=50\)
Vậy....