Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
ĐKXĐ: \(x\neq \pm 1\)
Ta có:
\(G=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right): \left(\frac{1}{x+1}-\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
\(=\frac{(x+1)^2-(x-1)^2}{(x+1)(x-1)}: \left(\frac{x-1}{(x+1)(x-1)}+\frac{x(x+1)}{(x-1)(x+1)}+\frac{2}{x^2-1}\right)\)
\(=\frac{x^2+2x+1-(x^2-2x+1)}{x^2-1}:\frac{x-1+x(x+1)+2}{x^2-1}\)
\(=\frac{4x}{x^2-1}:\frac{x^2+2x+1}{x^2-1}=\frac{4x}{x^2-1}.\frac{x^2-1}{(x+1)^2}=\frac{4x}{(x+1)^2}\)
Bài 4:
a: \(K=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{x+3}{x-2}+\dfrac{2x+1}{x-3}\)
\(=\dfrac{2x-9-x^2+9+\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{-x^2+2x+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để K<1 thì K-1<0
\(\Leftrightarrow\dfrac{x+1-x+3}{x-3}< 0\)
=>x-3<0
hay x<3
c: Để K là số nguyên thì \(x+1⋮x-3\)
\(\Leftrightarrow x-3+4⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{4;5;1;7;-1\right\}\)
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
mk nghỉ bài này đề sai
a) điều kiện : \(x\ne0;x\ne-1;x\ne2\)
ta có : \(A=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x-x^2-1}+\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(\Leftrightarrow A=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{1}{x^2-x+1}+\dfrac{2}{x+1}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{x+1+x+1+2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x\left(x-2\right)}{x^2-x+1}\) \(\Leftrightarrow A=1+\left(\dfrac{2x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2-x+1}{x\left(x-2\right)}\) \(\Leftrightarrow A=1+\dfrac{2x^2+4}{x\left(x+1\right)\left(x-2\right)}=\dfrac{2x^2+4+x\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)\left(x-2\right)}\)\(\Leftrightarrow A=\dfrac{x^3+x^2-2x+4}{x\left(x+1\right)\left(x-2\right)}\)
b) ta có : \(\left|x-\dfrac{3}{4}\right|=\dfrac{5}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{5}{4}\\x-\dfrac{3}{4}=\dfrac{-5}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(L\right)\\x=\dfrac{-1}{2}\end{matrix}\right.\)
thế vào \(A\) ta có : \(A=\dfrac{41}{5}\)
vậy ...............................................................................................................
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
a: \(=\dfrac{4a^2-4a+1-4a^2-2a+6a+3}{\left(2a-1\right)\left(2a+1\right)}\)
\(=\dfrac{4}{\left(2a-1\right)\left(2a+1\right)}\)
b: \(=\dfrac{x-1-x-1+2x^2}{\left(x-1\right)\left(x+1\right)}=2\)
d: \(=\dfrac{x-5+6x}{x\left(x+3\right)}=\dfrac{7x-5}{x\left(x+3\right)}\)
e: \(=\dfrac{x^2-4+3}{x-2}=\dfrac{x^2-1}{x-2}\)
i: \(=\dfrac{x}{x\left(x-4\right)}-\dfrac{3}{5x}=\dfrac{1}{x-4}-\dfrac{3}{5x}\)
\(=\dfrac{5x-3x+12}{5x\left(x-4\right)}=\dfrac{2x+12}{5x\left(x-4\right)}\)
a) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}=\dfrac{36}{x^2-9}\)
\(\Rightarrow\dfrac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\dfrac{\left(x+3\right)^2-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow\left(x+3\right)^2-\left(x-3\right)^2=36\)
\(\Rightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=36\)
\(\Rightarrow x^2+6x+9-x^2+6x-9=36\)
\(\Rightarrow12x=36\)
\(\Rightarrow x=\dfrac{36}{12}\)
Vậy x = 3
b) \(x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\)
\(\Rightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+17}{15}\)
\(\Rightarrow\dfrac{6x-3-5x+10}{15}=\dfrac{x+17}{15}\)
... Phần còn lại cũng tương tự như vậy thôi
a) \(G=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{1-x^2}\right):\left(\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}\right)\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{\left(1-x\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2+\left(x-1\right)\cdot\left(1-x\right)}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}+\dfrac{x}{-\left(x-1\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2+\left(x-1\right)\cdot\left(-\left(x-1\right)\right)}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\left(\dfrac{x+1}{x-1}+\dfrac{x}{x+1}-\dfrac{x}{\left(x-1\right)\cdot\left(1+x\right)}\right):\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x\cdot\left(x-1\right)-x}{\left(x-1\right)\cdot\left(x+1\right)}:\dfrac{2\cdot2x}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x^2-x-x}{\left(x-1\right)\cdot\left(x+1\right)}:\dfrac{4x}{\left(x-1\right)\cdot\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2+x^2-2x}{\left(x-1\right)\cdot\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{4x}\)
\(=\left(\left(x+1\right)^2+x^2-2x\right)\cdot\dfrac{1}{4x}\)
\(=\dfrac{\left(x+1\right)^2+x^2-2x}{4x}\)
\(=\dfrac{x^2+2x+1+x^2-2x}{4x}\)
\(=\dfrac{2x^2+1}{4x}\)