\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)

Lời giải:

$(x-5)(x+5)-(x+3)^2+3(x-2)^2=(x+1)^2-(x-4)(x+4)+3x^2$

$\Leftrightarrow x^2-25-(x^2+6x+9)+3(x^2-4x+4)=(x^2+2x+1)-(x^2-16)+3x^2$
$\Leftrightarrow 3x^2-18x-22=3x^2+2x+17$

$\Leftrightarrow -18x-22=2x+17$

$\Leftrightarrow 20x=-39$

$\Leftrightarrow x=\frac{-39}{20}$

\(\dfrac{3x-1}{3x^2+5x+2}=\dfrac{1}{x+2}\left(x\ne-2;x\ne\dfrac{1}{3}\right)\)

\(\Rightarrow\left(3x-1\right)\left(x+2\right)=3x^2+5x+2\)

\(\Rightarrow3x^2+6x-x-2=3x^2+5x+2\)

\(\Rightarrow3x^2+5x-2=3x^2+5x+2\)

\(\Rightarrow-2=2\) (vô lý)

Bạn xem lại đề bài

Câu khó thế

um khó mà giúp tui đi mn

\(\dfrac{x^2-3x}{2x^2-3x-9}=\dfrac{x^2+3x}{A}\)

\(\Rightarrow A=\dfrac{\left(x^2+3x\right)\left(2x^2-3x-9\right)}{x^2-3x}\)

\(\Rightarrow A=\dfrac{x\left(x+3\right)\left(2x^2-3x-9\right)}{x\left(x-3\right)}\)

\(\Rightarrow A=\dfrac{\left(x+3\right)\left(2x^2-3x-9\right)}{\left(x-3\right)}\)

mà \(x=-\dfrac{3}{2}\)

\(\Rightarrow A=\dfrac{\left(-\dfrac{3}{2}+3\right)\left(2\left(-\dfrac{3}{2}\right)^2-3\left(-\dfrac{3}{2}\right)-9\right)}{\left(-\dfrac{3}{2}-3\right)}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(2.\dfrac{9}{4}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}\)

\(\Rightarrow A=\dfrac{\dfrac{3}{2}\left(\dfrac{9}{2}+\dfrac{9}{2}-9\right)}{-\dfrac{9}{2}}=0\)

A = 0