Giải các bất phương trình sau:
a,1-3x/x-3 ≥1
b,1/x-5<1/x+2
c,2x/x-5 - 3/x-1<2
d,2x/x-5>1 + x-2/x+3
HELP ME, PLEASE
Giải các bất phương trình sau:
a,1-3x/x-3 ≥1
b,1/x-5<1/x+2
c,2x/x-5 - 3/x-1<2
d,2x/x-5>1 + x-2/x+3
HELP ME, PLEASE
Đáp án: Bài 1: Giải các bất phương trình sau:
a) $\frac{1-3x}{x-3}$ $\geq$ 1
ĐKXĐ: x $\neq$ 3
⇒ $\frac{1-3x}{x-3}$ – 1$\geq$ 0
⇒ $\frac{1-3x-(x-3)}{x-3}$$\geq$ 0
⇒ $\frac{4-4x}{x-3}$$\geq$ 0
⇒ $\left \{ {{4(x-1)\geq0 } \atop {x-3>0}} \right.$ ⇔ $\left \{ {{4(x-1)\leq0} \atop {x-3<0}} \right.$ $\frac{4(1-x)}{x-3}$$\geq$ 0
$\left \{ {{x\leq1} \atop {x>3}} \right.$ ⇔ $\left \{ {{x∈∅} \atop {x∈1;3}} \right.$
Vậy x ∈ {1;3}
b) $\frac{1}{x-5}$ < $\frac{1}{x+2}$
ĐKXĐ: x ≠ 1 và x≠ 5
$\frac{1}{x-5}$ < $\frac{1}{x+2}$
⇒ x – 5 > x + 2
⇒ 0x > 7
⇒ x = 0 ∈ ∅
c) $\frac{2x}{x-5}$ – $\frac{3}{x-1}$ < 2
ĐKXĐ: x≠ 5 và x≠ 1
$\frac{2x}{x-5}$ – $\frac{3}{x-1}$ < 2
⇒ $\frac{2x²-2x-3x+15}{(x-5)(x-1)}$ < 2
⇒ $\frac{2x²-5x+15-2(x²-6x+5)}{(x-5)(x-1)}$ < 0
⇒$\frac{7x-5}{(x-5)(x-1)}$ < 0
TH1) 7x – 5 > 0 ⇒ x > $\frac{5}{7}$
⇒ (x – 5)(x – 1)< 0
⇒ \(\left[ \begin{array}{l}\left \{ {{x-5>0} \atop {x-1<0}} \right. \\\left \{ {{x -5<0} \atop {x-1>0}} \right. \end{array} \right.\) ⇔ \(\left[ \begin{array}{l} \left \{ {{x>5} \atop {x<1}} \right. \\\left \{ {{x<5} \atop {x>1}} \right.\end{array} \right.\)
⇒ 1< x <5
+) TH2: 7x – 5 < 0 ⇒ x<$\frac{5}{7}$
⇒ (x – 5)(x – 1)>0
⇒\(\left[ \begin{array}{l}x>5\\x<1\end{array} \right.\)
⇒ x <$\frac{5}{7}$
Vậy \(\left[ \begin{array}{l}1<x<5\\x<\frac{5}{7}\end{array} \right.\)
d) $\frac{2x}{x-5}$ > 1+ $\frac{x-2}{x+3}$
ĐKXĐ: x ≠ -3 và x ≠ 5
$\frac{2x}{x-5}$ > 1+ $\frac{x-2}{x+3}$
⇒ $\frac{2x²+6x}{(x-5)(x+3)}$ >$\frac{x²-2x-15+x²-7x+10}{(x-5)(x+3)}$
⇒ $\frac{15x+5}{(x-5)(x+3)}$ > 0
⇒ $\frac{5(3x+1)}{(x-5)(x+3)}$ > 0
TH1) 3x + 1 > 0 ⇒ x > -$\frac{1}{3}$
⇒ (x – 5)(x+3) > 0
⇒ \(\left[ \begin{array}{l}x>5\\x<-3\end{array} \right.\)
⇒ x > 5
TH2) 3x + 1 < 0 ⇒ x < -$\frac{1}{3}$
⇒ (x – 5)(x + 3) < 0
⇒ -3 < x < 5
⇒ -3< x < -$\frac{1}{3}$
Vậy \(\left[ \begin{array}{l}x>5\\-3<x<-\frac{1}{3} \end{array} \right.\)
#Chúc bạn học tốt
Đáp án:
d. \(\left[ \begin{array}{l}
x > 5\\
– 3 < x < – \dfrac{1}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne 3\\
\dfrac{{1 – 3x}}{{x – 3}} \ge 1\\
\dfrac{{1 – 3x – x + 3}}{{x – 3}} \ge 0\\
\to \dfrac{{4 – 4x}}{{x – 3}} \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 – 4x \ge 0\\
x – 3 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
4 – 4x \le 0\\
x – 3 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 1\\
x > 3
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x \ge 1\\
x < 3
\end{array} \right.
\end{array} \right.\\
b.DK:x \ne \left\{ { – 2;5} \right\}\\
\dfrac{1}{{x – 5}} < \dfrac{1}{{x + 2}}\\
\to x – 5 > x + 2\\
\to 0x > 7\left( l \right)\\
\to x \in \emptyset \\
c.DK:x \ne \left\{ {1;5} \right\}\\
\dfrac{{2x}}{{x – 5}} – \dfrac{3}{{x – 1}} < 2\\
\to \dfrac{{2{x^2} – 2x – 3x + 15}}{{\left( {x – 5} \right)\left( {x – 1} \right)}} < 2\\
\to \dfrac{{2{x^2} – 5x + 15 – 2\left( {{x^2} – 6x + 5} \right)}}{{\left( {x – 5} \right)\left( {x – 1} \right)}} < 0\\
\to \dfrac{{7x – 5}}{{\left( {x – 5} \right)\left( {x – 1} \right)}} < 0\\
TH1:7x – 5 > 0 \to x > \dfrac{5}{7}\\
\to \left( {x – 5} \right)\left( {x – 1} \right) < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x – 5 > 0\\
x – 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x – 5 < 0\\
x – 1 > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 5\\
x < 1
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x < 5\\
x > 1
\end{array} \right.
\end{array} \right.\\
\to 1 < x < 5\\
TH2:7x – 5 < 0 \to x < \dfrac{5}{7}\\
\to \left( {x – 5} \right)\left( {x – 1} \right) > 0\\
\to \left[ \begin{array}{l}
x > 5\\
x < 1
\end{array} \right.\\
\to x < \dfrac{5}{7}\\
KL:\left[ \begin{array}{l}
1 < x < 5\\
x < \dfrac{5}{7}
\end{array} \right.\\
d.DK:x \ne \left\{ { – 3;5} \right\}\\
\dfrac{{2x}}{{x – 5}} > 1 + \dfrac{{x – 2}}{{x + 3}}\\
\to \dfrac{{2{x^2} + 6x}}{{\left( {x – 5} \right)\left( {x + 3} \right)}} > \dfrac{{{x^2} – 2x – 15 + {x^2} – 7x + 10}}{{\left( {x – 5} \right)\left( {x + 3} \right)}}\\
\to \dfrac{{15x + 5}}{{\left( {x – 5} \right)\left( {x + 3} \right)}} > 0\\
\to \dfrac{{5\left( {3x + 1} \right)}}{{\left( {x – 5} \right)\left( {x + 3} \right)}} > 0\\
TH1:3x + 1 > 0 \to x > – \dfrac{1}{3}\\
\to \left( {x – 5} \right)\left( {x + 3} \right) > 0\\
\to \left[ \begin{array}{l}
x > 5\\
x < – 3
\end{array} \right.\\
\to x > 5\\
TH2:3x + 1 < 0 \to x < – \dfrac{1}{3}\\
\to \left( {x – 5} \right)\left( {x + 3} \right) < 0\\
\to – 3 < x < 5\\
\to – 3 < x < – \dfrac{1}{3}\\
KL:\left[ \begin{array}{l}
x > 5\\
– 3 < x < – \dfrac{1}{3}
\end{array} \right.
\end{array}\)