Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)

\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)

ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)

\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)

\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)

Thỏa mãn ĐK

Các trường hợp khác làm tương tự

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)

\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)

\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)

\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)

\(a)=\frac{-2\left(x+3\right)}{x\left(1-3x\right)}.\frac{1-3x}{x\left(x+3\right)}\)

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

\(b)=\frac{\left(x+3\right)\left(x-3\right)}{x\left(x-3\right)}-\frac{x^2}{x\left(x-3\right)}+\frac{9}{x\left(x-3\right)}\)

\(=\frac{x^2-3x+3x-9-x^2+9}{x\left(x-3\right)}\)

\(=x\left(x-3\right)\)

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

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

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

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

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

# Sắp ik ngủ nên làm vậy hoi, ko chắc phần kq câu b và c đâu nha

a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)

\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)

\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)

\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)

\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)

\(\Leftrightarrow-25x=-13\)

\(\Leftrightarrow x=\dfrac{13}{25}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)

gắp cái gì

Lần sau đăng thì chia thành nhiều câu hỏi nhé

\(16^2-9.\left(x+1\right)^2=0\)

\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)

\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)

\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)

\(\left[13-3x\right].\left[19+3x\right]=0\)

\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)

KL:..............................

Nhiều câu hỏi mà bn ??

       \(2x-2=8-3x\)

\(\Leftrightarrow\)\(2x+3x=8+2\)

\(\Leftrightarrow\)\(5x=10\)

\(\Leftrightarrow\)\(x=2\)

Vậy...

         \(x^2-3x+1=x+x^2\)

\(\Leftrightarrow\)\(x^2-3x-x-x^2=-1\)

\(\Leftrightarrow\)\(-4x=-1\)

\(\Leftrightarrow\)\(x=\frac{1}{4}\)

Vậy...

mấy cái này bấm máy tính là đc òi. giải mất thời gian lắm :))

a. \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{9}{x^2-9}\) (ĐKXĐ: \(x\ne\pm3\))

\(\Leftrightarrow\left(x+3\right)^2-\left(x-3\right)^2=9\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=9\)

\(\Leftrightarrow12x=9\Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

\(\Rightarrow S=\left\{\dfrac{3}{4}\right\}\)

b. \(\dfrac{x+2}{4}-x+3=\dfrac{1-x}{8}\)

\(\Leftrightarrow2\left(x+2\right)-8\left(x-3\right)=1-x\)

\(\Leftrightarrow2x+4-8x+24=1-x\)

\(\Leftrightarrow2x-8x+x=1-4-24\)

\(\Leftrightarrow-3x=-27\Leftrightarrow x=9\)

\(\Rightarrow S=\left\{9\right\}\)

-Mệt -.-