Question

[Trang Huyen]

giúp tớ với các cou vừa nãy câu hỏi bị lỗi nên xin lỗi các cou nha

Answer 1

[Akai Haruma]

Bài 1:

a) $2(x-2)-3(x-2)=0$

$\Leftrightarrow -(x-2)=0$

$\Leftrightarrow x=2$

b) ĐK: $x\neq \pm 3$

$\frac{5}{x-3}+\frac{4}{x+3}=\frac{x-5}{x^2-9}$

$\Leftrightarrow \frac{5(x+3)+4(x-3)}{x^2-9}=\frac{x-5}{x^2-9}$

$\Leftrightarrow \frac{9x+3}{x^2-9}=\frac{x-5}{x^2-9}$
$\Rightarrow 9x+3=x-5$

$8x=-8$

$x=-1$ (thỏa)

c) 

$(5x+7)(10-5x)=0$

$\Leftrightarrow 5x+7=0$ hoặc $10-5x=0$

$\Leftrightarrow x=-\frac{7}{5}$ hoặc $x=2$

d)

$|2x-3|=5-x$ 

\(\Rightarrow \left\{\begin{matrix} 5-x\geq 0\\ \left[\begin{matrix} 2x-3=5-x\\ 2x-3=x-5\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{8}{3}\end{matrix}\right.\)

Answer 2

[Akai Haruma]

Bài 2:

a)

BPT \(\Leftrightarrow \frac{5(4x-1)}{15}-\frac{2-x}{15}\leq \frac{3(10x-3)}{15}\)

\(\Leftrightarrow 5(4x-1)-(2-x)\leq 3(10x-3)\)

\(\Leftrightarrow 9x\geq 2\Leftrightarrow x\geq \frac{2}{9}\)

b) 

\(\frac{1-2x}{4}-2\leq \frac{1-5x}{8}+x+2\Leftrightarrow \frac{1-2x-8}{4}\leq \frac{3x+17}{8}\)

\(\Leftrightarrow 2(-2x-7)\leq 3x+17\)

\(\Leftrightarrow 7x\geq -31\Leftrightarrow x\geq \frac{-31}{7}\)