Câu 1 :
a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)
b.
|-2x|=5x+14
Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14
<=> - 2x = 5x + 14
<=> - 2x - 5x = 14
<=> - 7x = 14
<=> x = - 2 (thoã mãn)
Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.
Ta có pt: 2x = 5x + 14
<=> - 3x = 14
<=> x = \(-\dfrac{14}{3}\)
Vậy pt có nghiệm x = - 2
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)
\(a,4x-5=23\)
\(\Leftrightarrow4x=23+5\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(b,\left|-2x\right|=5x+14\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)
\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)
\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)
\(\Leftrightarrow x=0\)
Vậy \(S=\left\{0\right\}\)
a) \(4x-5=23\)
\(4x=23+5\)
\(4x=28\)
\(x=7\)
b) \(\left|-2x\right|=5x+14\)
\(\Leftrightarrow\) \(-2x-5=14\)
\(\Leftrightarrow\) \(-7x=14\)
\(\Leftrightarrow\) \(x=-2\)
\(\Leftrightarrow\) \(-2x=-\left(5x+14\right)\)
\(\Leftrightarrow\) \(-2x=-\left(5x-14\right)\)
\(\Leftrightarrow\) \(-2x+5x=-14\)
\(\Leftrightarrow\) \(3x=-14\)
\(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)
\(\Leftrightarrow x=-2\)
c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x-1}+\dfrac{-1}{x+1}=\dfrac{x^2+2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)+\left(-1\right)\left(x-1\right)=x^2+2\)
\(\Leftrightarrow x^2+x+2=x^2+2\)
\(\Leftrightarrow x+2=2\)
\(\Leftrightarrow x=0\)
\(a,4x-5=23\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
\(b,\left|-2x\right|=5x+14\)
*TH1 : \(\left|-2x\right|=2x\Leftrightarrow x>0\)
\(\Rightarrow2x=5x+14\)
\(\Leftrightarrow2x-5x=14\)
\(\Leftrightarrow-3x=14\)
\(\Leftrightarrow x=-\dfrac{14}{3}\left(loại\right)\)
*TH2 : \(\left|-2x\right|=-2x\Leftrightarrow x< 0\)
\(\Rightarrow-2x=5x+14\)
\(\Leftrightarrow-2x-5x=14\)
\(\Leftrightarrow-7x=14\)
\(\Leftrightarrow x=-2\left(nhận\right)\)
Vậy \(S=\left\{-2\right\}\)
\(c,\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\left(đkxđ:x\ne\pm1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)-x+1=x^2+2\)
\(\Leftrightarrow x^2+x+x+1-x+1=x^2+2\)
\(\Leftrightarrow x^2-x^2+x+x-x+1+1-2=0\)
\(\Leftrightarrow x=0\left(nhận\right)\)
Vậy \(S=\left\{0\right\}\)
