\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
a. Rút gọn P
b. Tìm x sao cho P<1
c.
tìm \(x\in Z\)để \(P\in Z\)
Giúp mk nha mn! Mk đang cần gấp!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ĐKXĐ : \(x\ge0,x\ne25,x\ne9\)
a) \(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\frac{x-5\sqrt{x}-\left(x-25\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right):\left(\frac{-\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}:\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=-\frac{5}{\sqrt{x}+5}:\frac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}=\frac{-5}{\sqrt{x}+5}.\left(\frac{-\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\right)=\frac{5}{\sqrt{x}+3}\)
b) \(A< 1\Rightarrow\frac{5}{\sqrt{x}+3}< 1\Rightarrow\sqrt{x}+3>5\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
Chú ý kết hợp với điều kiện xác định.
Sửa đề: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
Ta có: \(\dfrac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}-\dfrac{\sqrt{x}+5}{\sqrt{x}-3}-\dfrac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\dfrac{-x+8\sqrt{x}-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}+\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-31-\left(x-25\right)+3x-9\sqrt{x}-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{2x-2\sqrt{x}-28-x+25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-2\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x-3\sqrt{x}+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)
Câu 3 :
\(ĐKXĐ:x>0\)
\(P=\left(\frac{2}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+2}\right):\frac{2\sqrt{x}}{x+2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}\cdot\frac{x+2\sqrt{x}}{2\sqrt{x}}\)
\(\Leftrightarrow P=\frac{2\sqrt{x}+4+x}{2\sqrt{x}}\)
b) Để P = 3
\(\Leftrightarrow\frac{2\sqrt{x}+4+x}{x+2\sqrt{x}}=3\)
\(\Leftrightarrow2\sqrt{x}+4+x=6\sqrt{x}\)
\(\Leftrightarrow x-4\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
\(\Leftrightarrow\sqrt{x}-2=0\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)(tm)
Vậy để \(P=3\Leftrightarrow x=4\)
Câu 1 : Hình như sai đề !! Mik sửa :
\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
\(A=\left(\frac{x}{x\sqrt{x}-4\sqrt{x}}-\frac{6}{3\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right):\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{6}{\sqrt{x}+2}\)
\(\Leftrightarrow A=\frac{-6\left(\sqrt{x}+2\right)}{6\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(\Leftrightarrow A=-\frac{1}{\sqrt{x}-2}\)
b) Để A < 2
\(\Leftrightarrow-\frac{1}{\sqrt{x}-2}< 2\)
\(\Leftrightarrow-1< 2\sqrt{x}-4\)
\(\Leftrightarrow2\sqrt{x}>3\)
\(\Leftrightarrow\sqrt{x}>1,5\)
\(\Leftrightarrow x>2,25\)
Vậy để \(A< 2\Leftrightarrow x>2,25\)
\(a,\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\x\ne9;x\ne25\end{cases}}\)
\(P=\frac{8\sqrt{x}-x-31}{x-8\sqrt{x}+15}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}-\frac{3\sqrt{x}-1}{5-\sqrt{x}}\)
\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(-\frac{\sqrt{x}+15}{\sqrt{x}-3}+\frac{3\sqrt{x}-1}{\sqrt{x}-5}\)
\(=\frac{8\sqrt{x}-x-31}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}-\)\(\frac{\left(\sqrt{x}+15\right)\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)\(+\frac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{8\sqrt{x}-x-31-x-10\sqrt{x}+75+3x-10\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x-12\sqrt{x}+47}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)}\)
\(\Rightarrow\)Sai đề không cậu ưi