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12 tháng 2 2020
a) ĐKXĐ: x≠-5
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow\frac{2x-5}{x+5}-3=0\)
\(\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)
\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)
\(\Leftrightarrow2x-5-3x-15=0\)
\(\Leftrightarrow-x-20=0\)
\(\Leftrightarrow-x=20\)
\(\Leftrightarrow x=-20\)(tmđk)
Vậy: x=-20
b) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2}{x-1}=\frac{6}{x+1}\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0\)
\(\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow2x+2-6x+6=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)(tmđk)
Vậy: x=2
c) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+x+12x-4=0\)
\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)
\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(thỏa mãn điều kiện)
Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)
d) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x\left(x+1\right)-2x=0\)
\(\Leftrightarrow x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy: x=0
e) ĐKXĐ: x≠2
Ta có: \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
⇔\(\frac{1}{x-2}+3-\frac{x-3}{2-x}=0\)
⇔\(\frac{1}{x-2}+3+\frac{x-3}{x-2}=0\)
⇔\(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)
\(\Leftrightarrow1+3\left(x-2\right)+x-3=0\)
\(\Leftrightarrow1+3x-6+x-3=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)(không thỏa mãn)
Vậy: x∈∅
f) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
⇔\(\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0\)
⇔\(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0\)
\(\Leftrightarrow0=0\)
Vậy: x∈R
g) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+2}{x-2}+\frac{1}{x+2}=\frac{x\left(x-5\right)}{x^2-4}\)
⇔\(\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\left(x+2\right)^2+x-2-x^2+5x=0\)
\(\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow10x=2\)
\(\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}\)(thỏa mãn)
Vậy: \(x=\frac{1}{5}\)
18 tháng 8 2021
Áp dụng bất đẳng thức AM-GM ta có :
\(B=\frac{12}{x-1}+\frac{x-1+1}{3}=\frac{12}{x-1}+\frac{x-1}{3}+\frac{1}{3}\ge2\sqrt{\frac{12}{x-1}\cdot\frac{x-1}{3}}+\frac{1}{3}=4+\frac{1}{3}=\frac{13}{3}\)
Dấu "=" xảy ra <=> \(\frac{12}{x-1}=\frac{x-1}{3}\Rightarrow x=7\left(x\ge1\right)\). Vậy MinB = 13/3
HN
16 tháng 12 2018
\(a,\frac{2x+4}{10}+\frac{2-x}{15}=\frac{\left(2x+4\right).3}{10.3}+\frac{\left(2-x\right).2}{15.2}\)
\(=\frac{6x+12}{30}+\frac{4-2x}{30}=\frac{6x+12+4-2x}{30}=\frac{4x+16}{30}\)
\(=\frac{4.\left(x+4\right)}{30}=\frac{2\left(x+4\right)}{15}\)
\(b,\frac{3x}{10}+\frac{2x-1}{15}+\frac{2-x}{20}=\frac{3x.6}{10.6}+\frac{\left(2x-1\right).4}{15.4}+\frac{\left(2-x\right).3}{20.3}\)
\(=\frac{18x}{60}+\frac{8x-4}{60}+\frac{6-3x}{60}=\frac{18x+8x-4+6-3x}{60}=\frac{23x+2}{60}\)
\(c,\frac{x+1}{2x-2}+\frac{x^2+3}{2-2x^2}=\frac{x+1}{2\left(x-1\right)}+\frac{x^2+3}{2\left(1-x^2\right)}=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x^2-1\right)}\)
\(=\frac{x+1}{2\left(x-1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\frac{-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2-3}{2\left(x-1\right)\left(x+1\right)}=\frac{2x-2}{2\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\)\(=\frac{1}{x+1}\)
MT
1
16 tháng 2 2017
c) \(\frac{x-3}{x-2}+\frac{x-2}{x-4}=1\) đặt x-2 =t " cho bé hệ số lại
ĐK : \(\left\{\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\Rightarrow\left\{\begin{matrix}t\ne0\\t\ne-2\end{matrix}\right.\)
\(\frac{t-1}{t}=\frac{t}{t-2}\Leftrightarrow\left(t-1\right)\left(t-2\right)=t^2\Leftrightarrow t^2-3t+2=t^2\Rightarrow-3t=-2\)
\(t=\frac{2}{3}\Rightarrow x=2+\frac{2}{3}=\frac{8}{3}\)
20 tháng 2 2017
a) \(A=\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3x}\) xem lại đề thấy cái mẫu VP vô duyên thế!
b) \(B=\frac{2}{x-1}+\frac{2x+3}{x^2+x+1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}\) MSC=(x^3-1)
\(B=\frac{2\left(x^2+x+1\right)+\left(2x+3\right)\left(x-1\right)-\left(4x^2-1\right)}{MSC}=\frac{\left(2x^2+2x+2\right)+\left(2x^2+x-3\right)-4x^2+1}{MSC}=0\)
\(B=0\Leftrightarrow\frac{3x}{MSC}=0=>x=0\) thảo mãn đk x khác 1
Kết luận: x=0 là nghiệm duy nhất.