Bài 1: Tìm số nguyên x để thỏa mãn phương trình:
a/ $\frac{3x-2}{5}$ ≥ $\frac{x}{2}$ +0,8 và 1- $\frac{2x-5}{6}$ > $\frac{3-x}{4}$
b/ 2(3x-4) < 3(4x-3)+16 và a(1+x) < 3x+5
Bài 2: Cho biểu thức:
A= ( $\frac{1}{1-x}$ + $\frac{2}{x+1}$ - $\frac{5-x}{1-x mũ2}$) : $\frac{1-2x}{x mũ2-1}$
a) Rút gọn
b) Tìm x để A>0
Đáp án:
$\begin{array}{l}
1)a)\dfrac{{3x – 2}}{5} \ge \dfrac{x}{2} + 0,8\\
\Rightarrow \dfrac{3}{5}.x – \dfrac{2}{5} \ge \dfrac{x}{2} + \dfrac{4}{5}\\
\Rightarrow \dfrac{3}{5}.x – \dfrac{x}{2} \ge \dfrac{4}{5} + \dfrac{2}{5}\\
\Rightarrow \dfrac{1}{{10}}.x \ge \dfrac{6}{5}\\
\Rightarrow x \ge 12\\
Khi:1 – \dfrac{{2x – 5}}{6} > \dfrac{{3 – x}}{4}\\
\Rightarrow 1 – \dfrac{x}{3} + \dfrac{5}{6} > \dfrac{3}{4} – \dfrac{x}{4}\\
\Rightarrow \dfrac{x}{3} – \dfrac{x}{4} < 1 + \dfrac{5}{6} – \dfrac{3}{4}\\
\Rightarrow \dfrac{x}{{12}} < \dfrac{{13}}{{12}}\\
\Rightarrow x < 13\\
\Rightarrow 12 \le x < 13\\
\Leftrightarrow x = 12\\
Vậy\,x = 12\\
b)2\left( {3x – 4} \right) < 3\left( {4x – 3} \right) + 16\\
\Leftrightarrow 6x – 8 < 12x – 9 + 16\\
\Leftrightarrow 12x – 6x > – 15\\
\Rightarrow x > \dfrac{{ – 5}}{2}\\
a.\left( {1 + x} \right) < 3x + 5\\
\Rightarrow a + a.x < 3x + 5\\
\Rightarrow \left( {a – 3} \right).x < 5 – a\\
B2)\\
a)Dkxd:x \ne 1;x \ne – 1\\
A = \left( {\dfrac{1}{{1 – x}} + \dfrac{2}{{x + 1}} – \dfrac{{5 – x}}{{1 – {x^2}}}} \right):\dfrac{{1 – 2x}}{{{x^2} – 1}}\\
= \dfrac{{ – x – 1 + 2\left( {x – 1} \right) + 5 – x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x – 1} \right)\left( {x + 1} \right)}}{{1 – 2x}}\\
= \dfrac{{ – x – 1 + 2x – 2 + 5 – x}}{{1 – 2x}}\\
= \dfrac{2}{{1 – 2x}}\\
b)x \ne 1;x \ne – 1\\
A > 0\\
\Leftrightarrow \dfrac{2}{{1 – 2x}} > 0\\
\Leftrightarrow 1 – 2x > 0\\
\Leftrightarrow x < \dfrac{1}{2}\\
Vậy\,x < \dfrac{1}{2};x \ne – 1
\end{array}$