Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
Đề sai ạ ! Sửa nhé :
\(S=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^3-4x}{2x^2-x^3}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)}{x-2}+\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(x^2-4\right)}{x^2\left(2-x\right)}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)}{-x\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x+2\right)\left(x-2\right)}.\frac{-x}{\left(x+2\right)}\)
\(\Leftrightarrow S=\frac{-x\left(4x^2-8x\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2}{\left(x+2\right)^2}\)
P/s : nếu làm theo đề của bạn, sẽ ra kq dài... Nên mik tiện sửa, còn nếu đề bạn đúng rồi thì mik sẽ làm lại ạ !
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)\(\Leftrightarrow\frac{x^2+3x+2+x^2-3x+2}{x^2-4}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow2\left(x^2+2\right)=2\left(x^2+2\right)\)(luôn đúng)
Vậy pt có vô số nghiệm
\(b,\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)\(\Leftrightarrow\left(\frac{-4x+10}{2-7x}\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+10=0\\x+8=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}\)
Mấy câu rút gọn bạn quy đồng nha
Điều kiện: x\(\ne\) 0; x \(\ne\) 2; -2; 3
A=\(\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{2+x}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
A = \(\left(\frac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right).\frac{x\left(2-x\right)}{\left(x-3\right)}\)
A = \(\frac{x^2+4x+4+4x^2-\left(4-4x+x^2\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{\left(x-3\right)}\)
A = \(\frac{8x+4x^2}{\left(2+x\right)}.\frac{x}{\left(x-3\right)}=\frac{4x\left(x+2\right)}{\left(x+2\right)}.\frac{x}{x-3}=\frac{4x^2}{x-3}\)