\(2\left(5x+1\right)-5=\left(x+6\right)\left(2x+k\right)\left(1\right)\)

Thay \(x=3\) vào \(\left(1\right)\)ta có:

\(2.\left(5.3+1\right)-5=\left(3+6\right)\left(2.3+k\right)\)

Thu gọn, tìm được \(k=-3\)

= 0.2 chứ làm sao

:)))

:))))))

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`Answer:`

\(\frac{x-2}{x-3}=\frac{x-3+1}{x-3}=1+\frac{1}{x-3}\)

\(\Rightarrow x-3\inƯ\left(1\right)=\left\{\pm1\right\}\)

cảm ơn bạn

\(-2x^2-10x-6=0\)\(\Leftrightarrow4x^2+20x+12=0\)\(\Leftrightarrow\left(4x^2+20x+25\right)-13=0\)\(\Leftrightarrow\left(2x+5\right)^2-\left(\sqrt{13}\right)^2=0\)\(\Leftrightarrow\left(2x+5+\sqrt{13}\right)\left(2x+5-\sqrt{13}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x+5+\sqrt{13}=0\\2x+5-\sqrt{13}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5-\sqrt{13}}{2}\\x=\frac{-5+\sqrt{13}}{2}\end{cases}}\)

\(PT\Leftrightarrow2x^2+10x+6=0\)

\(\Leftrightarrow x^2+5x+3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{5+\sqrt{13}}{2}\\x=-\frac{5-\sqrt{13}}{2}\end{cases}}\)

-2x-10x-6=0

<=>-12x=6

<=>x=\(\frac{6}{-12}\)

<=>x=\(\frac{-1}{2}\)

Vậy x\(\in\left\{\frac{-1}{2}\right\}\)

x=-1/2

\(A=\frac{x^2-x}{x^2-9}-\frac{1}{x-3}+\frac{1}{x+3}\)

\(=\frac{x\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}-\frac{1}{x-3}+\frac{1}{x+3}\)

\(=\frac{x\left(x-1\right)-x-3+x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x^2+x-6}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{\left(x-2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{x-2}{x-3}\)

\(A=\frac{x^2-x}{x^2-9}-\frac{1}{x-3}+\frac{1}{x+3}\)

\(A=\frac{x^2-x}{x^2-3^2}-\frac{1}{x-3}+\frac{1}{x+3}\)

\(A=\frac{x^2-x}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x-3}{\left(x-3\right)\left(x-3\right)}\)

\(A=\frac{x^2-x-\left(x+3\right)+x-3}{\left(x-3\right)\left(x+3\right)}\)

\(A=\frac{x^2-x-x-3+x-3}{\left(x-3\right)\left(x+3\right)}\)

\(A=\frac{x^2-x}{\left(x-3\right)\left(x+3\right)}\)

 (3x-1)(x+3)= (2-x)(5-3x) 

\(\Leftrightarrow3x^2+9x-x-3=10-6x-5x+3x^2\)

\(\Leftrightarrow3x^2+8x-3-10+11x-3x^2=0\)

\(\Leftrightarrow19x-13=0\)

\(\Leftrightarrow x=\frac{13}{19}\)

Vậy \(x\in\left\{\frac{13}{19}\right\}\)

hình như sai đề á mk lm k ra mk nghĩ là sai th