a. \(5x-10=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b. \(\dfrac{2}{x+1}-\dfrac{3}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne-1;x\ne2\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{3\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2\left(x-2\right)-3\left(x+1\right)=2x-6\)
\(\Leftrightarrow2x-4-3x-3=2x-6\)
\(\Leftrightarrow2x-4-3x-3-2x+6=0\)
\(\Leftrightarrow-3x-1=0\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
c) \(3x-5\ge-7\) (3)
\(\Leftrightarrow3x\ge-7+5\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
Vậy tập nghiệm của BPT (3) là \(x\ge-\dfrac{2}{3}\)
d) \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(S=\left\{\dfrac{1}{3}\right\}\)
a. 5x - 10 = 0
⇔ 5x = 10
⇔ x = 2
Vậy S ={2}.
b.\(\dfrac{2}{x+1}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔\(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\) - \(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔ 2(x - 2) - 3(x+1) = 2x - 6
⇔ 2x - 4 - 3x -3 = 2x - 6
⇔ 2x - 2x - 3x = -6 + 4 + 3
⇔ -3x = 1
⇔ x = \(-\dfrac{1}{3}\)
a. 5x-10=0
<=>5x=10
<=>x=2
Vậy phương trình có nghiệm là x=2
b.\(\dfrac{2}{x+1}\)-\(\dfrac{3}{x-2}\)=\(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\).ĐKXĐ:x\(\ne\)1;x\(\ne\)2
<=> \(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
=>2x-4-3x-3=2x-6
<=>-x-7=2x-6
<=>-x-2x=-6+7
<=>-3x=1
<=>x=\(\dfrac{-1}{3}\)
Vậy phương trình có nghiệm là x=\(\dfrac{-1}{3}\)
c.3x-5\(\ge\)-7
<=>3x\(\ge\)-2
<=>x\(\ge\)\(\dfrac{-2}{3}\)
Vậy bất phương trình có nghiệm là x\(\ge\)\(\dfrac{-2}{3}\)
d.3x-1=0
<=>3x=1
<=>x=\(\dfrac{1}{3}\)
Vậy phương trình có nghiệm là x=\(\dfrac{1}{3}\)
a. 5x - 10 = 0
⇒5x = 10
⇒x = 2
b. \(\dfrac{2}{x+1}-\dfrac{3}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇒\(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇒2x - 4 - 3x - 1 = 2x - 6
⇒3x = 1
⇒x = \(\dfrac{1}{3}\)
c. 3x − 5 ≥ − 7
⇒3x ≥ -2
⇒x ≥ \(\dfrac{-2}{3}\)
d. 3x - 1 = 0
⇒3x = 1
⇒x = \(\dfrac{1}{3}\)
