Question

\(a,\left(3x-2\right)\left(x+1\right)=x^2-1\)

\(b,\dfrac{2x-5}{x+4}+\dfrac{x}{4-x}=\dfrac{-17x+56}{x^2-16}\)

\(c,\dfrac{3x+5}{4}-\dfrac{x-4}{6}\le\dfrac{3x+7}{3}\)

Answer 1

Lần sau bạn lưu ý ghi đầy đủ đề bài. 

Giải PT/ HPT:

a. $(3x-2)(x+1)=x^2-1$

$\Leftrightarrow (3x-2)(x+1)-(x^2-1)=0$

$\Leftrightarrow (3x-2)(x+1)-(x+1)(x-1)=0$

$\Leftrightarrow (x+1)[(3x-2)-(x-1)]=0$

$\Leftrightarrow (x+1)(2x-1)=0$

$\Leftrightarrow x+1=0$ hoặc $2x-1=0$

$\Leftrightarrow x=-1$ hoặc $x=\frac{1}{2}$

b. ĐKXĐ: $x\neq \pm 4$

$\frac{2x-5}{x+4}+\frac{x}{4-x}=\frac{-17x+56}{x^2-16}$

$\Leftrightarrow \frac{(2x-5)(x-4)-x(x+4)}{(x-4)(x+4)}=\frac{-17x+56}{x^2-16}$

$\Leftrightarrow \frac{x^2-17x+20}{x^2-16}=\frac{-17x+56}{x^2-16}$

$\Rightarrow x^2-17x+20=-17x+56$

$\Leftrightarrow x^2=36$

$\Leftrightarrow x=\pm 6$ (tm)

c/

$\frac{3x+5}{4}-\frac{x-4}{6}\leq \frac{3x+7}{3}$

$\Leftrightarrow \frac{3(3x+5)-2(x-4)}{12}\leq \frac{4(3x+7)}{12}$
$\Leftrightarrow 3(3x+5)-2(x-4)\leq 4(3x+7)$

$\Leftrightarrow -5x-5\leq 0$

$\Leftrightarrow -5x\leq 5\Leftrightarrow x\geq -1$