giair pt sau :
d) 3/x+2-x+2/x-2 = 5x-2/4-x ²-1
e) /x-5/-9=3x
f) x(x-2)=3(x-2)
giair pt sau : d) 3/x+2-x+2/x-2 = 5x-2/4-x ²-1 e) /x-5/-9=3x f) x(x-2)=3(x-2)
By Natalia
By Natalia
giair pt sau :
d) 3/x+2-x+2/x-2 = 5x-2/4-x ²-1
e) /x-5/-9=3x
f) x(x-2)=3(x-2)
Đáp án:
$\begin{array}{l}
d)Dkxd:x \ne 2;x \ne – 2\\
\dfrac{3}{{x + 2}} – \dfrac{{x + 2}}{{x – 2}} = \dfrac{{5x – 2}}{{4 – {x^2}}} – 1\\
\Rightarrow \dfrac{{3\left( {x – 2} \right) – {{\left( {x + 2} \right)}^2}}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{{5x – 2}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} + 1 = 0\\
\Rightarrow \dfrac{{3x – 6 – {x^2} – 4x – 4 + 5x – 2 + {x^2} – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = 0\\
\Rightarrow 4x – 16 = 0\\
\Rightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4\\
e)\left| {x – 5} \right| – 9 = 3x\\
\Rightarrow \left| {x – 5} \right| = 3x + 9\left( {dk:3x + 9 \ge 0 \Rightarrow x \ge – 3} \right)\\
\Rightarrow \left[ \begin{array}{l}
x – 5 = 3x + 9\\
x – 5 = – 3x – 9
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = – 7\left( {ktm} \right)\\
x = – 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = – 1\\
f)x\left( {x – 2} \right) = 3\left( {x – 2} \right)\\
\Rightarrow \left( {x – 2} \right)\left( {x – 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 3
\end{array} \right.
\end{array}$
`d, \frac{3}{x+2} – \frac{x+2}{x-2} = \frac{5x-2}{4-x^2} – 1` $\text{(đk: x $\neq$ ± 2)}$
`⇔ \frac{3}{x+2} – \frac{x+2}{x-2} = – \frac{5x-2}{x^2-4} – 1`
`⇔ \frac{3(x-2)-(x+2)^2+5x-2+(x-2)(x+2)}{(x-2)(x+2)} = 0`
`⇒ 3(x-2) – (x+2)^2 + 5x – 2 + (x-2)(x+2)=0`
`⇔ 3x – 6 – x^2 – 4x – 4 + 5x – 2 + x^2 -4 = 0`
`⇔ 4x = 16`
`⇔ x = 4` `(TM)`
$\text{Vậy tập nghiệm của PT: S = {4}}$
`e. |x-5| – 9 = 3x`
$\text{Có: |x-5| = x – 5 nếu x – 5 ≥ 0 ⇔ x ≥ 5}$
$\text{|x-5| = -(x – 5) = 5 – x nếu x – 5 < 0 ⇔ x < 5}$
$\text{Với x ≥ 5}$
`⇒ x – 5 – 9 = 3x`
`⇔ -2x = 14`
`⇔ x = -7` `(KTM)`
$\text{Với x < 5}$
`⇒ 5 – x – 9 = 3x`
`⇔ -4x = 4`
`⇔ x = -1` `(TM)`
$\text{Vậy tập nghiệm của PT: S = {-1}}$
`f, x(x-2) = 3(x-2)`
`⇔ x^2 – 2x = 3x – 6`
`⇔ x^2 -5x + 6 = 0`
`⇔ x^2 – 3x – 2x + 6 = 0`
`⇔ x(x – 3) – 2(x – 3) = 0`
`⇔ (x – 2)(x – 3) = 0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\x-3=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\)
$\text{Vậy tập nghiệm của PT: S = {2; 3}}$