𝑎)2𝑥−1𝑥−3+4=−1𝑥−3
⇔2x-1x+1x=-3+3-4
⇔2x=-4
⇔x=-2
𝑏)3𝑥−22𝑥+5=6𝑥+14𝑥−3
⇔5+3=6x+14x-3x+22x
⇔8=39x
⇔x=\(\frac{8}{39}\)
𝑐)𝑥+3𝑥+1+𝑥−2𝑥=2
⇔x+3x+x-2x=2-1
⇔3x=1
⇔x=\(\frac{1}{3}\)
𝑑)x+1−2𝑥−3𝑥−1=2𝑥+3𝑥2−1
⇔3x2+2x+2x+3x-x-1-1+1=0
⇔3x2+6x-1=0
⇔3x2+3x+3x+3-4=0
⇔3x(x+1)+3(x+1)-4=0
⇔3(x+1)(x+1)-4=0
⇔3(x+1)2-4=0
⇔(x+1)2=\(\frac{4}{3}\)
⇔\(\left[{}\begin{matrix}x+1=\frac{4}{3}\\x+1=-\frac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}-1\\x=-\frac{4}{3}-1\end{matrix}\right.\)
Vậy ...
a, 2x - x - 3 + 4 = -x - 3
\(\Leftrightarrow\) x + 1 = -x - 3
\(\Leftrightarrow\) x + x = -3 - 1
\(\Leftrightarrow\) 2x = -4
\(\Leftrightarrow\) x = -2
Vậy S = {-2}
b, 3x - 22x + 5 = 6x + 14x - 3
\(\Leftrightarrow\) -19x + 5 = 20x - 3
\(\Leftrightarrow\) -19x - 20x = -3 - 5
\(\Leftrightarrow\) -39x = -8
\(\Leftrightarrow\) x = \(\frac{8}{39}\)
Vậy S = {\(\frac{8}{39}\)}
c, x + 3x + 1 + x - 2x = 2
\(\Leftrightarrow\) 3x + 1 = 2
\(\Leftrightarrow\) 3x = 2 - 1
\(\Leftrightarrow\) 3x = 1
\(\Leftrightarrow\) x = \(\frac{1}{3}\)
Vậy S = {\(\frac{1}{3}\)}
Phần d mình ko hiểu, bạn viết rõ được ko!
Chúc bn học tốt!!
d, x + 1 - 2x - 3x - 1 = 2x + 3x2 - 1
\(\Leftrightarrow\) x + 1 - 2x - 3x - 1 - 2x - 3x2 + 1 = 0
\(\Leftrightarrow\) -3x2 - 6x + 1 = 0
\(\Leftrightarrow\) -(3x2 + 6x - 1) = 0
\(\Leftrightarrow\) 3x2 + 6x - 1 = 0
\(\Leftrightarrow\) 3x2 + 3x + 3x + 3 - 4 = 0
\(\Leftrightarrow\) 3x(x + 1) + 3(x + 1) - 4 = 0
\(\Leftrightarrow\) 3(x + 1)(x + 1) - 4 = 0
\(\Leftrightarrow\) 3(x + 1)2 - 4 = 0
\(\Leftrightarrow\) (x + 1)2 = \(\frac{4}{3}\)
\(\Leftrightarrow\) x + 1 = \(\sqrt{\frac{4}{3}}\) hoặc x + 1 = \(-\sqrt{\frac{4}{3}}\)
\(\Leftrightarrow\) x = \(\sqrt{\frac{4}{3}}\) - 1 và x = \(-\sqrt{\frac{4}{3}}\) - 1
\(\Leftrightarrow\) x = \(\frac{2\sqrt{3}-3}{3}\) và x = \(\frac{-2\sqrt{3}-3}{3}\)
Vậy S = {\(\frac{2\sqrt{3}-3}{3}\); \(\frac{-2\sqrt{3}-3}{3}\)}
Chúc bn học tốt!!
