a/ \(2x-3=5x+2\)
\(\Leftrightarrow5x-2x=-3-2\)
\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy..
b. \(2x\left(x-1\right)=2x+2\)
\(\Leftrightarrow2x^2-4x-2=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1+\sqrt{2}\right)\left(x-1-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)
Vậy...
c/ ĐKXĐ : \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)
\(\Leftrightarrow2x-16=0\)
\(\Leftrightarrow x=8\)
Vậy..
a/ \(2x-3=5x+2\)
\(< =>2x-5x=3+2\)
\(< =>-3x=5\)
\(< =>x=-\dfrac{5}{3}\)
Vậy.....
b/ ( đàu bài thiếu)
c/ \(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{12}{2\left(x+2\right)}\)
ĐKXĐ của phương tình là: \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{12}{2\left(x+2\right)}\)
\(< =>\dfrac{2\left(x+2\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}-\dfrac{2x^2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{12\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(< =>\left(2x+4\right)\left(x+2\right)-2x^2-12x+24=0\)
\(< =>2x^2+4x+4x+8-2x^2-12x+24=0\)
\(< =>-4x+32=0\)
\(< =>-4x=-32\)
\(< =>x=-32:-4=7\)
Vậy....
a) 2x - 3 = 5x + 2
⇔ 2x - 5x = 3 + 2
⇔ -3x = 5
⇔ x = \(\dfrac{-5}{3}\)
Vậy PT có tập nghiệm S = \(\left\{\dfrac{-5}{3}\right\}\)
b) 2x( x - 1 ) - 2x + 2 = 0
⇔ 2x ( x - 1 ) - 2( x - 1 ) = 0
⇔ ( 2x - 2 )( x - 1 ) = 0
⇒ 2( x - 1 )( x - 1 ) = 0
⇒ x - 1 = 0
⇔ x = 1
Vậy PT có tập nghiệm S = { 1 }
c)\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{12}{2\left(x+2\right)}\) ( ĐKXĐ : x ≠ 2 và x ≠ -2 )
⇔ \(\dfrac{2\left(x+2\right)\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{2x^2}{2\left(x+2\right)\left(x-2\right)}=\dfrac{12\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}\)
⇔ 2x2 + 8x + 8 -2x2 = 12x - 24
⇔ 8x - 12x = -8 - 24
⇔ -4x = -32
⇔ x = 8 ( thoả mãn ) d
Vậy PT có tập nghiệm S = { 8 }