PP chung ở cả 3 câu,nói ngắn gọn nhé:

Chứng mình x khác 0,hay nói cách khác x=0 không là nghiệm của phương trình.

Chia cả tử và mẫu cho x ,rồi giải bình thường bằng cách đặt ẩn phụ.

Vd ở câu a>>>4/(4x-8+7/x)+3/(4x-10+7/x)=1.Sau đó đặt 4x+7/x=a>>>4/(a-8)+3/(a-10)=1>>>giải bình thường,các câu sau tương tự

a,\(\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}\)

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

\(=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{\left(2x-1\right).2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{2.2\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}\)

\(=\frac{3x-3}{2x\left(x+1\right)\left(x-1\right)}+\frac{4x^2-2x}{2x\left(x-1\right)\left(x+1\right)}-\frac{4x^2-4}{2x\left(x+1\right)\left(x-1\right)}\)

\(=\frac{3x-3+4x^2-2x-4x^2+4}{2x\left(x+1\right)\left(x-1\right)}\)

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

\(b,\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

\(=\frac{3x.2\left(x-y\right)}{10\left(x+y\right).\left(x-y\right)}-\frac{x.\left(x+y\right)}{10\left(x-y\right).\left(x+y\right)}\)

\(=\frac{6x^2-6xy}{10\left(x+y\right)\left(x-y\right)}-\frac{x^2+xy}{10\left(x-y\right)\left(x+y\right)}\)

\(=\frac{6x^2-6xy-x^2+xy}{10\left(x+y\right)\left(x-y\right)}\)

\(=\frac{5x^2-5xy}{10\left(x+y\right)\left(x+y\right)}\)

\(=\frac{5x\left(x-y\right)}{10\left(x-y\right)\left(x+y\right)}=\frac{x}{2\left(x+y\right)}\)

\(a)\)  Ta có : 

\(\frac{x}{18}=\frac{y}{9}\)\(\Leftrightarrow\)\(\frac{x}{2}=y\)

\(\Rightarrow\)\(x=2y\)

Thay \(x=2y\) vào \(A=\frac{2x-3y}{2x+3y}\) ta được : 

\(A=\frac{2.2y-3y}{2.2y+3y}=\frac{4y-3y}{4y+3y}=\frac{y}{7y}=\frac{1}{7}\)

Vậy ... ( tự kết luận ) 

Chúc bạn học tốt ~ 

ỳgyjwegfeukwfhưe

phùng minh quân giải mình câu b,c,d vs đc ko bạn

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Ta nhân tích chéo lên bạn nhé!

 => ( 2x + 3 )( 10x + 2) = ( 5x+2)(4x+5)

  <=> 20x² + 34x + 6 = 20x² + 33x + 10

    <=> x = 4   ( chỗ này chuyễn vế sang bạn nhé)

Sau đó kết luận thôi>>>>.......................

a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)

= 2x(x+y) - y(x+y)  /  2x(x-y) - y(x-y)

= (2x-y)(x+y)  /  (2x-y)(x-y)

= x+y/x-y

Rút gọn cái sau:

\(\frac{32x+4x^2+2x^3}{x^3+64}\)

\(=\frac{2x\left(x^2+2x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)

Đề có vẻ sai sai ?