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\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne30\\x\ne24\end{cases}}\)
Ta có \(\frac{60}{\frac{120}{x}-4}+\frac{60}{\frac{120}{x}-5}=x\)
\(\Leftrightarrow\frac{60}{\frac{120-4x}{x}}+\frac{60}{\frac{120-5x}{x}}=x\)
\(\Leftrightarrow\frac{60x}{120-4x}+\frac{60x}{120-5x}=x\)
\(\Leftrightarrow\frac{60}{120-4x}+\frac{60}{120-5x}=1\left(Do\text{ }x\ne0\right)\)
\(\Leftrightarrow\frac{15}{30-x}=1-\frac{12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{24-x-12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{12-x}{24-x}\)
\(\Leftrightarrow360-15x=\left(12-x\right)\left(30-x\right)\)
\(\Leftrightarrow360-15x=360-42x+x^2\)
\(\Leftrightarrow x^2-27x=0\)
\(\Leftrightarrow x\left(x-27\right)=0\)
\(\Leftrightarrow x=27\left(Tm\text{ }ĐKXĐ\right)\)
a,12x-180+10x-20+39x-2340+65x-4420=780
126x-6960=780
126x=7740
x=430/7
3/(x^2-13x+40)+2/(x^2-8x+15)+1/(x^2-5x+6)+6/5+0
3/(x-8)(x-5)+2/(x-5)(x-3)+1/(x-3)(x-2)+6/5=0
1/(x-8)-1/(x-5)+1/(x-5)-1/(x-3)+1/(x-3)-1/(x-2)+6/5=0
1/(x-8)-1/(x-2)+6/5=0
ban tu giai tiep nhan
m^2x+2x=5-3mx
m^2x+3mx+2x=5
x(m^2+3m+2)=5
khi 0x=5 thi pt vo nghiem
m^2+3m+2=0
(m+1)(m+2)=0
m=-1 hoac m=-2
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
điều kiện: \(x\ne5;8\)
\(\frac{6\left(x-8\right)+2\left(x-5\right)}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-13x+40}+1=0\)
\(\frac{6x-48+2x-10}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x^2-8x-5x+40}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{x\left(x-8\right)-5\left(x-8\right)}+1=0\)
\(\frac{8x-58}{\left(x-5\right)\left(x-8\right)}-\frac{18}{\left(x-5\right)\left(x-8\right)}+\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{8x-58-18+x^2-13x+40}{\left(x-5\right)\left(x-8\right)}=0\)
\(\frac{x^2-5x-36}{\left(x-5\right)\left(x-8\right)}=0\)
=> \(x^2-5x-36=0\)
\(x^2+4x-9x-36=0\)
\(x\left(x+4\right)-9\left(x+4\right)=0\)
\(\left(x-9\right)\left(x+4\right)=0\)
Vậy x - 9 = 0 hoặc x + 4 = 0
hay x = 9 (thỏa mãn điều kiện) hoặc x = -4 (thỏa mãn điều kiện)
vậy...
\(\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{x^2-13x+40}-1\)
ĐKXĐ: \(x\ne5,8\)
\(\Leftrightarrow\frac{6}{x-5}+\frac{2}{x-8}=\frac{18}{\left(x-5\right)\left(x-8\right)}-1\)
\(\Leftrightarrow6\left(x-8\right)+2\left(x-5\right)=18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow8x-58=-22-x^2+13x\)
\(\Leftrightarrow8x-58+22+x^2-13x=0\)
\(\Leftrightarrow-5x-36+x^2=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
Vậy: phương trình có tập nghiệm là: S = {9; -4}
\(ĐKXĐ:x\ne\pm5\)
\(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5}{x+5}+\frac{x-3}{x-5}=\frac{2x-40}{x^2-25}\)
\(\Leftrightarrow\frac{5x-25+5x-15}{x^2-25}=\frac{2x-40}{x^2-25}\)
\(\Rightarrow10x-40=2x-40\)
\(\Leftrightarrow x=0\left(TMĐKXĐ\right)\)
Vậy x=0
\(\frac{5}{x+5}-\frac{x-3}{5-x}=\frac{2x-40}{x^2-25}\) ( đkxđ : \(x\ne\pm5\))
( 5 - x ) = -( 5 - x ) = -5 + x = x - 5
<=> \(\frac{5}{x+5}-\frac{x-3}{x-5}=\frac{2x-40}{\left(x+5\right)\left(x-5\right)}\)
<=> \(\frac{5\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}-\frac{\left(x+5\right)\left(x-3\right)}{\left(x+5\right)\left(x-5\right)}=\frac{2x-40}{\left(x+5\right)\left(x-5\right)}\)
<=> \(5x-25-x^2+2x-15=2x-40\)
<=> \(5x-x^2+2x-2x=-40+25+15\)
<=> \(5x-x^2=0\)
<=> \(x^2-5x=0\)
<=> \(x\left(x-5\right)=0\)
<=> x = 0 ( nhận ) hoặc x = 5 ( loại do đkxđ )
Vậy nghiệm của phương trình là x = 0
<=>
\(5x-200=\frac{5x}{2}-300+2x+300\)0
\(3x-2,5x=200\)\(0,5x=200\)\(x=400\)