\(\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne1;x\ne-1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)+\left(1-x\right)\left(x-1\right)=4\)

\(\Leftrightarrow x^2+x+x+1+x-1-x^2+x=4\)

\(\Leftrightarrow x^2-x^2+x+x+x+x-4+1-1=0\)

\(\Leftrightarrow4x-4=0\)

\(\Leftrightarrow4x=4\)

\(\Leftrightarrow x=1\left(loại\right)\)

Vậy \(S=\varnothing\)

  \(\dfrac{x+1}{x-1}+\dfrac{1-x}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(1-x\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{\left(x+1+x-1\right)\left(x+1-x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{2x.2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇔ \(\dfrac{4x}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

⇒ \(4x=4\)

⇔ \(x=1\) (ko thỏa mãn dktc)

Vậy phương trình vô nghiệm

cíu mình đi ng ơi huhu

a: ĐKXĐ: x-2>=0 và 6-2x>=0

=>2<=x<=3

b: DKXĐ: x+2>=0

=>x>=-2

a)\(=\dfrac{x-2+x+5}{x+3}=\dfrac{2x+3}{x+3}\)

b)ĐKXĐ: \(x+3\ne0\Leftrightarrow x\ne-3\)

a) Ta có: \(\dfrac{x-2}{x+3}+\dfrac{x+5}{x+3}\)

\(=\dfrac{x-2+\left(x+5\right)}{x+3}\)

\(=\dfrac{x-2+x+5}{x+3}\)

\(=\dfrac{2x+3}{x+3}\)

b) ĐKXĐ: \(x+3\ne0\)

hay \(x\ne-3\)

bt xác định \(\Leftrightarrow\sqrt{x-1}-2\ne0\)

\(\Leftrightarrow\sqrt{x-1}\ne2\)

\(\Leftrightarrow x-1\ne4\Leftrightarrow x\ne5\)

Bạn Đào Trọng Luân thiếu ĐK căn lớn hơn bằng 0

\(ĐKXĐ\hept{\begin{cases}x-1\ge0\\\sqrt{x-1}\ne2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x-1\ne4\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\ne5\end{cases}}}\)

-Không có ĐKXĐ, tức là với mọi x∈R thì phân thức luôn được xác định.

a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{9;4\right\}\end{matrix}\right.\)

b: Ta có: \(P=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(=\dfrac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

mấy bạn vô câu hỏi tương tự nhe