Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(-ĐKXĐ:\hept{\begin{cases}x-14\ne0;x-13\ne0\\x-9\ne0\\x-11\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne14;x\ne13\\x\ne9\\x\ne11\end{cases}}\)
- Ta có : \(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)
\(\Leftrightarrow\frac{2}{x-14}-\frac{5}{x-13}-\frac{2}{x-9}+\frac{5}{x-11}=0\)
\(\Leftrightarrow\left(\frac{2}{x-14}-\frac{2}{x-9}\right)-\left(\frac{5}{x-13}-\frac{5}{x-11}\right)=0\)
\(\Leftrightarrow2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)-5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)=0\)\(\Leftrightarrow2.\frac{\left(x-9\right)-\left(x-14\right)}{\left(x-9\right)\left(x-14\right)}-5.\frac{\left(x-11\right)-\left(x-13\right)}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow2.\frac{5}{\left(x-9\right)\left(x-14\right)}-5.\frac{2}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow\frac{10}{\left(x-9\right)\left(x-14\right)}-\frac{10}{\left(x-11\right)\left(x-13\right)}=0\)
\(\Leftrightarrow10\left[\frac{1}{\left(x-9\right)\left(x-14\right)}-\frac{1}{\left(x-11\right)\left(x-13\right)}\right]=0\)
\(\Leftrightarrow\frac{\left(x-11\right)\left(x-13\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}-\frac{\left(x-9\right)\left(x-14\right)}{\left(x-9\right)\left(x-14\right)\left(x-11\right)\left(x-13\right)}=\) \(0\)
\(\Leftrightarrow\left(x-11\right)\left(x-13\right)-\left(x-9\right)\left(x-14\right)=0\)
\(\Leftrightarrow x^2-24x+143-x^2+23x-126=0\)
\(\Leftrightarrow-x+17=0\Leftrightarrow-x=-17\Leftrightarrow x=17\)
Vậy pt có tập nghiệm S = { 17 }
P/s: Mk làm hơi lòng vòng, bn thông cảm nhé !
a) ĐKXĐ: \(x\notin\pm\frac{1}{3}\)
Ta có: \(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{9\left(12x-4x^2-1\right)}{4\left(9x^2-1\right)}\)
\(\Leftrightarrow\frac{2\left(12x+1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{4\left(9x-5\right)\left(3x-1\right)}{4\left(3x+1\right)\left(3x-1\right)}=\frac{9\left(12x-4x^2-1\right)}{4\left(3x+1\right)\left(3x-1\right)}\)
\(\Leftrightarrow72x^2+30x+2-\left(108x^2-96x+20\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+30x+2-108x^2+96x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow9\left(2x-1\right)=0\)
mà 9≠0
nên 2x-1=0
⇔2x=1
hay \(x=\frac{1}{2}\)(tm)
Vậy: \(x=\frac{1}{2}\)
b)ĐKXĐ: x≠0
Ta có: \(x+\frac{1}{x}=x^2+\frac{1}{x^2}\)
\(\Leftrightarrow x+\frac{1}{x}-x^2-\frac{1}{x^2}=0\)
\(\Leftrightarrow\frac{x^3}{x^2}+\frac{x}{x^2}-\frac{x^4}{x^2}-\frac{1}{x^2}=0\)
\(\Leftrightarrow x^3+x-x^4-1=0\)
\(\Leftrightarrow x^3\left(1-x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow x^3\left(1-x\right)-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow-\left(x-1\right)^2\cdot\left(x^2+x+1\right)=0\)(1)
Ta có: \(x^2+x+1=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)(2)
Từ (1) và (2) suy ra x-1=0
hay x=1(tm)
Vậy: x=1
c) ĐKXĐ: x≠0
Ta có: \(\frac{1}{x}+2=\left(\frac{1}{x}+2\right)\left(x^2+2\right)\)
\(\Leftrightarrow\frac{1}{x}+2-\left(\frac{1}{x}+2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\left(2-x^2-2\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}+2\right)\cdot\left(-x^2\right)=0\)(3)
Ta có: 1≠0
x≠0
Do đó: \(\frac{1}{x}\ne0\)
\(\Leftrightarrow\frac{1}{x}+2\ne0\)(4)
Từ (3) và (4) suy ra x=0(ktm)
Vậy: x∈∅
d) ĐKXĐ: x≠0
Ta có: \(\left(x+1+\frac{1}{x}\right)^2=\left(x-1-\frac{1}{x}\right)^2\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}\right)^2-\left(x-1-\frac{1}{x}\right)^2=0\)
\(\Leftrightarrow\left(x+1+\frac{1}{x}+x-1-\frac{1}{x}\right)\left(x+1+\frac{1}{x}-x+1+\frac{1}{x}\right)=0\)
\(\Leftrightarrow2x\cdot\left(2+\frac{2}{x}\right)=0\)
\(\Leftrightarrow4x\left(1+\frac{1}{x}\right)=0\)
mà 4≠0
và x≠0
nên \(1+\frac{1}{x}=0\)
\(\Leftrightarrow\frac{1}{x}=-1\)
hay x=-1(tm)
Vậy: x=-1
\(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đkxđ \(x\ne\pm\frac{1}{3}\)
\(\Leftrightarrow\frac{12x+1}{2\left(3x-1\right)}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow\frac{\left(24x+2\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}-\frac{\left(36x-20\right)\left(3x-1\right)\left(3x+1\right)}{4\left(3x-1\right)\left(3x+1\right)}=\frac{-36x^2+10x-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Leftrightarrow72x^2+6x+24x+2-108x^2+60x+36x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\left(tmđk\right)\)
\(\frac{12x+1}{6x-2}-\frac{9x-5}{3x+1}=\frac{108x-36x^2-9}{4\left(9x^2-1\right)}\)đkxđ \(x\ne\pm\frac{1}{3}\)
\(\Leftrightarrow72x^2+6x+24x+2-108x^2+60x+36x-20-108x+36x^2+9=0\)
\(\Leftrightarrow18x-9=0\)
\(\Leftrightarrow18x=9\)
\(\Leftrightarrow x=\frac{1}{2}\left(tm\right)\)
a) \(\frac{x+\frac{2\left(3-x\right)}{5}}{14}-\frac{5x-4\left(x-1\right)}{24}=\frac{7x+2+\frac{9-3x}{5}}{12}+\frac{2}{3}\)
<=> \(\frac{5x+2\left(3-x\right)}{70}-\frac{5x-4\left(x-1\right)}{24}=\frac{35x+10+9-3x}{60}+\frac{2}{3}\)
<=> \(12\left(5x+6-2x\right)-35\left(5x-4x+4\right)\)
<=> \(14\left(35x+10+9-3x\right)+280.2\) <=> \(12\left(3x+6\right)-35\left(x+4\right)\)
<=> \(14\left(32x+19\right)+560\)
<=> \(36x+72-35x-140=448x+226+560\)
<=> \(-447x=894\)
<=> x = -2
a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)
\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)
\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)
\(\Leftrightarrow46x-429=0\)
\(\Leftrightarrow46x=429\)
hay \(x=\frac{429}{46}\)
Vậy: \(x=\frac{429}{46}\)
b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)
\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)
\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)
\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)
\(\Leftrightarrow-685x+261.5=0\)
\(\Leftrightarrow-685x=-261.5\)
hay \(x=\frac{523}{1370}\)
Vậy: \(x=\frac{523}{1370}\)
c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)
\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)
\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)
\(\Leftrightarrow-125x+423=0\)
\(\Leftrightarrow-125x=-423\)
hay \(x=\frac{423}{125}\)
Vậy: \(x=\frac{423}{125}\)
d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x+20x-140=0\)
\(\Leftrightarrow-37x+259=0\)
\(\Leftrightarrow-37x=-259\)
hay \(x=7\)
Vậy: x=7
