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\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
\(ĐK:x\ne0;-10\)
\(\Leftrightarrow\dfrac{12\left(x+10\right)+12x}{12x\left(x+10\right)}=\dfrac{x\left(x+10\right)}{12x\left(x+10\right)}\)
\(\Leftrightarrow12\left(x+10\right)+12x-x\left(x+10\right)=0\)
\(\Leftrightarrow12x+120+12x-x^2-10x=0\)
\(\Leftrightarrow-x^2+14x+120=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)
\(o,\dfrac{x}{2x+6}-\dfrac{x}{2x-2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x+3\right)}-\dfrac{x}{2\left(x+1\right)}-\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)-x\left(x+3\right)-2\left(3x+2\right)}{2\left(x+1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+x-x^2-3x-6x-4=0\)
\(\Leftrightarrow-8x-4=0\)
\(\Leftrightarrow-4\left(2x+1\right)=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)
a) Quy đồng bỏ mẫu rồi giai pt ta đc : \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
b)\(x=1\)
1) Ta có: \(4x+8=3x-1\)
\(\Leftrightarrow4x-3x=-1-8\)
\(\Leftrightarrow x=-9\)
2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)
\(\Leftrightarrow10-5x-15-3x+3>0\)
\(\Leftrightarrow-8x>2\)
hay \(x< \dfrac{-1}{4}\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\left(ĐKXĐ:x\ne-10;x\ne0\right)\)
\(\Leftrightarrow\frac{12\left(x+10\right)}{12x\left(x+10\right)}+\frac{12x}{12x\left(x+10\right)}=\frac{x\left(x+10\right)}{12x\left(x+10\right)}\)
\(\Rightarrow12\left(x+10\right)+12x=x\left(x+10\right)\)
\(\Leftrightarrow12x+120+12x=x^2+10x\)
\(\Leftrightarrow x^2+10x-12x-12x-120=0\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow\left(x-20\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-20=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=20\\x=-6\end{cases}}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm \(S=\left\{-6;20\right\}\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
ĐKXĐ : x khác 0 ; x khác -10
<=> \(\frac{x+10}{x\left(x+10\right)}+\frac{x}{x\left(x+10\right)}=\frac{1}{12}\)
<=> \(\frac{2x+10}{x\left(x+10\right)}=\frac{1}{12}\)
=> 24x + 120 = x2 + 10x
<=> x2 + 10x - 24x - 120 = 0
<=> x2 - 14x - 120 = 0
<=> x2 - 20x + 6x - 120 = 0
<=> x( x - 20 ) + 6( x - 20 ) = 0
<=> ( x - 20 )( x + 6 ) = 0
<=> x = 20 hoặc x = -6 ( tm )
Vậy S = { 20 ; -6 }