c/ \(\dfrac{x+4}{x+1}-2=\dfrac{2-x}{x}\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
a) \(5-\left(x-6\right)=4\left(3-2x\right)\\ < =>5-x+6=12-8x\\ < =>-x+8x=5+6-12\\ < =>7x=-1\\ =>x=-\dfrac{1}{7}\)
Vậy: Tập nghiệm của pt là S= {-1/7}.
b, (3x+1)(3x-2)= (5x-8)(3x+1)
<=> (3x+1)(3x-2)-(5x-8)(3x+1)= 0
<=> (3x+1)(3x-2-5x+8)=0
<=> (3x+1)(6-2x)=0
<=> 3x+1=0 hoặc 6-2x=0
* 3x+1=0 <=> 3x=-1 <=> x= -1/3
* 6-2x=0 <=> 2x=6 <=> x=3
Vậy tập nghiệm của pt là S=[ -1/3; 3]
b)
\(\left(3x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(3x+1\right)\\ \Leftrightarrow\left(3x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(3x+1\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(3x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-2x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{6}{-2}=3\end{matrix}\right.\)
Vây phương trình có tập nghiệm là S={-1/3; 3}
a) 5-(x-6)=4(3-2x)
\(\Leftrightarrow\) 5-x+6=12-8x
\(\Leftrightarrow\)8x-x=12-5-6
\(\Leftrightarrow\)7x=1
\(\Leftrightarrow\)x=\(\dfrac{1}{7}\)
b)\(\left(3x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(3x+1\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\6-2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=3\end{matrix}\right.\)
c)\(\dfrac{x+4}{x+1}-2=\dfrac{2-x}{x}\)
(ĐKXĐ: \(x\ne0;x\ne\left(-1\right)\))
\(\Rightarrow\dfrac{x^2+4x}{x\left(x+1\right)}-\dfrac{2x^2+2x}{x\left(x+1\right)}=\dfrac{\left(2-x\right)\left(x+1\right)}{x\left(x+1\right)}\)
\(\Rightarrow x^2+4x-2x^2-2x=-x^2+x+2\)
\(\Leftrightarrow x^2+4x-2x^2-2x+x^2-x=2\)
\(\Leftrightarrow2x=2\)
\(\Leftrightarrow x=1\)(thõa)
