Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 : Điều kiện xác định : \(x\ne\pm1\)
\(K=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x^2-1}{x^2}\)
\(K=\frac{2}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x^2}=\frac{2}{x^2}\)
Nhận thấy giá trị của x càng tăng thì giá trị của M càng giảm
mặt khác , giá trị của x lại không giảm quá 0 nên ta không thể nào xác định được giá trị lớn nhất của K
\(a,ĐKXĐ:x\ne\sqrt{2};-\sqrt{2};x\ne4\)
\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{x-4}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}+\frac{-2-5\sqrt{x}}{x-4}\)
\(P=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)
\(P=\frac{3x-6\sqrt{x}}{x-4}\)
\(b;\)Để P<2
\(\Rightarrow3x-6\sqrt{x}< 2x-8\)
\(\Rightarrow3x-2x< -8+6\sqrt{x}\)
\(\Rightarrow x-6\sqrt{x}< -8\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-6\right)< 8\)
Tìm x là xong
a) \(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x>4\right)\)
\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(P=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
b) Ta có : \(P< 2\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}< 2\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-2< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{\sqrt{x}+2}-\frac{2\sqrt{x}+4}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}+2}< 0\)
Mà \(\sqrt{x}-4< \sqrt{x}+2\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}+2>0\end{cases}\Leftrightarrow}\hept{\begin{cases}\sqrt{x}< 4\\\sqrt{x}>-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 16\\x>4\end{cases}}\Leftrightarrow4< x< 16\)
Vậy ...
a)ĐKXĐ:
\(x-1\ne0;x+1\ne0;x\ne0\)
\(\Leftrightarrow x\ne1;x\ne-1;x\ne0\)
b)\(K=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2003}{x}\)
\(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)
\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{x+2003}{x}\)
\(=\frac{x^2+2x+1+x^2-2x+1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x^2-4x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x^2-3x-x+1}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{3x.\left(x-1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(x+1\right)}.\frac{x+2003}{x}\)
\(=\frac{\left(3x-1\right)\left(x+2003\right)}{\left(x+1\right).x}\)
\(=\frac{3x^2+6008x-2003}{x^2+x}\)
câu c bí
Bài làm
a) \(Q=\left(\frac{1}{x-1}-\frac{1}{x}\right):\left(\frac{x+1}{x-2}-\frac{x+2}{x-1}\right)\)
\(Q=\left(\frac{x}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}\right):\left(\frac{x^2-1}{\left(x-2\right)\left(x-1\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-1\right)}\right)\)
\(Q=\left(\frac{x-x+1}{x\left(x-1\right)}\right):\left(\frac{x^2-1-x^2+4}{\left(x-2\right)\left(x-1\right)}\right)\)
\(Q=\frac{1}{x\left(x-1\right)}:\frac{3}{\left(x-2\right)\left(x-1\right)}\)
\(Q=\frac{1}{x\left(x-1\right)}.\frac{\left(x-2\right)\left(x-1\right)}{3}\)
\(Q=\frac{x-2}{3x}\)
ĐKXĐ: \(\frac{x-2}{3}\ge0\)
Vì \(\frac{x-2}{3}\ge0\). Mà 3 > 0
=> x - 2 > 0
<=> x > 2
Vậy x > 2 thì biểu thức Q có nghĩa.
b) \(C=\left(\frac{x+2}{x^2-x}+\frac{x-2}{x^2+x}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{x+2}{x\left(x-1\right)}+\frac{x-2}{x\left(x+1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{\left(x+2\right)\left(x+1\right)}{x\left(x^2-1\right)}+\frac{\left(x-2\right)\left(x-1\right)}{x\left(x^2-1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\left(\frac{x^2+2x+x+2+x^2-x-2x+2}{x\left(x^2-1\right)}\right).\frac{x^2-1}{x^2+2}\)
\(C=\frac{2x^2+4}{x\left(x^2-1\right)}.\frac{x^2-1}{x^2+2}\)
\(C=\frac{2\left(x^2+2\right)}{x\left(x^2-1\right)}.\frac{x^2-1}{x^2+2}\)
\(C=\frac{2}{x}\)
ĐKXĐ: \(\frac{2}{x}\ge0\)
Vì \(\frac{2}{x}\ge0\),
Mà 2 > 0
=> x > 0
Vậy x > 0 thì biểu thức C có nghĩa.