a)\(3x-2=2x+5\\ \Leftrightarrow x=7\\ \Rightarrow S=\left\{7\right\}\)
b)\(2\left(x+3\right)< 4x-\left(2+x\right)\)
\(\Leftrightarrow2x+6< 4x-2-x\\ \Leftrightarrow-x< -8\\ \Leftrightarrow x>8\)
\(S=\left\{x\in R|x>8\right\}\)
c)\(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow6x^2-x-2=10x^2-11x-8\\ \Leftrightarrow4x^2-10x-6=0\\ \Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow S=\left\{-\dfrac{1}{2};3\right\}\)
d)\(\dfrac{x+6}{5}-\dfrac{x-2}{3}< 2\)
\(\Leftrightarrow\dfrac{3x+18-5x+10-30}{15}< 0\\ \Leftrightarrow\dfrac{-2x-2}{15}< 0\\ \Leftrightarrow-2x-2< 0\left(15>0\right)\\ \Leftrightarrow x>-1\)
Vậy \(S=\left\{x\in R|x>-1\right\}\)
e)\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\left(x\ne2;x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)-2\left(x-11\right)}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x+2\right)\left(x-2\right)}=0\\ \Leftrightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x-5\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\\ \Leftrightarrow S=\left\{4;5\right\}\)
Tick hộ nha 😘
a) \(3x-2=2x+5\)
\(\Leftrightarrow x=7\)
b) \(2\left(x+3\right)< 4x-\left(2+x\right)\)
\(\Leftrightarrow2x+6< 4x-2-x\)
\(\Leftrightarrow2x+6< 3x-2\)
\(\Leftrightarrow-x< -8\)
\(\Leftrightarrow x>8\)
c) \(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
d) \(\dfrac{x+6}{5}-\dfrac{x-2}{3}< 2\)
\(\Leftrightarrow\dfrac{3\left(x+6\right)-5\left(x-2\right)-30}{15}< 0\)
\(\Leftrightarrow\dfrac{3x+18-5x+10-30}{15}< 0\)
\(\Leftrightarrow\dfrac{-2-2x}{15}< 0\)
Vì 15 > 0 nên để \(\dfrac{-2-2x}{15}< 0\) thì \(-2-2x< 0\) \(\Leftrightarrow x>-1\)
e) ĐKXĐ: \(x\ne2;x\ne-2\)
Với \(x\ne2;x\ne-2\), ta có:
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
\(\Leftrightarrow\dfrac{\left(x-2\right)^2-3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2x-22}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4x+4-3x-6=2x-22\)
\(\Leftrightarrow x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
a) 3x - 2 = 2x + 5
\(\Leftrightarrow\) 3x - 2 - 2x - 5 = 0
\(\Leftrightarrow\) x - 7 = 0
\(\Leftrightarrow\) x = 7
Vậy S = \(\left\{7\right\}\)
b) 2(x + 3) < 4x - (2 + x)
\(\Leftrightarrow\) 2x + 6 < 4x - 2 - x
\(\Leftrightarrow\) 2x - 4x + x < -2 - 6
\(\Leftrightarrow\) -x < -8
\(\Leftrightarrow\) x < 8
c) (2x + 1)(3x - 2) = (5x - 8)(2x + 1)
\(\Leftrightarrow\) (2x + 1)(3x - 2) - (5x - 8)(2x + 1) = 0
\(\Leftrightarrow\) (2x - 1)\(\left[\left(3x-2\right)-\left(5x-8\right)\right]\) = 0
\(\Leftrightarrow\) (2x - 1)(3x - 2 - 5x + 8) = 0
\(\Leftrightarrow\) (2x - 1)(-2x + 6) = 0
Khi 2x - 1 = 0 hoặc -2x + 6 = 0
\(\Leftrightarrow\) 2x = 1 \(\Leftrightarrow\) -2x = -6
\(\Leftrightarrow\) x = \(\dfrac{1}{2}\) \(\Leftrightarrow\) x = \(\dfrac{-6}{-2}=3\)
Vậy S = \(\left\{\dfrac{1}{2};3\right\}\)
d) \(\dfrac{x+6}{5}-\dfrac{x-2}{3}< 2\)
MTC : 15
⇒ \(\dfrac{3\left(x+6\right)}{15}-\dfrac{5\left(x-2\right)}{15}< \dfrac{30}{15}\)
Suy ra : 3(x + 6) - 5(x - 2) < 30
\(\Leftrightarrow\) 3x + 18 - 5x + 10 < 30
\(\Leftrightarrow\) 3x - 5x < 30 - 18 - 10
\(\Leftrightarrow\) -2x < 2
\(\Leftrightarrow\) x >\(\dfrac{2}{-2}=-1\)
e) \(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
dkxd : x ≠ -2
x ≠ 2
MTC : (x - 2)(x + 2)
Quy đồng mãu thức hai vế của phương trình :
⇒ \(\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\) \(\dfrac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra : (x - 2)(x - 2) - 3(x + 2) = 2(x - 11)
\(\Leftrightarrow\) (x - 2)2 - 3(x + 2) = 2(x - 11)
\(\Leftrightarrow\) x2 - 4x + 4 - 3x - 6 = 2x - 22
\(\Leftrightarrow\) x2 - 4x + 4 - 3x - 6 - 2x + 22 = 0
\(\Leftrightarrow\) x2 -9x + 20 = 0
\(\Leftrightarrow\) x2 - 4x - 5x + 20 = 0
\(\Leftrightarrow\) x(x - 4) - 5 (x - 4) = 0
\(\Leftrightarrow\) (x - 4)(x - 5) = 0
Khi x - 4 = 0 hoặc x - 5 = 0
\(\Leftrightarrow\) x = 4 \(\Leftrightarrow\) x = 5
Vậy S = \(\left\{4;5\right\}\)
a) Ta có: 3x-2=2x+5
nên 3x-2x=5+2
hay x=7
b) Ta có: \(2\left(x+3\right)< 4x-\left(2+x\right)\)
\(\Leftrightarrow2x+6-4x+x+2< 0\)
\(\Leftrightarrow-x< -8\)
hay x>8
c: Ta có: \(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
d: Ta có: \(\dfrac{x+6}{5}-\dfrac{x-2}{3}< 2\)
\(\Leftrightarrow\dfrac{3x+18}{15}-\dfrac{5x-10}{15}< \dfrac{30}{15}\)
\(\Leftrightarrow3x+18-5x+10-30< 0\)
\(\Leftrightarrow-2x< 2\)
hay x>-1
