Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Chào em, em có thể kam khảo tại link:
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
Nếu link bị chặn em copy và dán tại:
https://olm.vn/hoi-dap/question/1261852.html
Câu hỏi của Lê Thu Hà - Toán lớp 9 - Học toán với OnlineMath
a) Rút gọn E
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}\div\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-\sqrt{x}}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}-\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\div\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
Vậy \(E=\frac{x}{\sqrt{x}-1}\)
a ) \(ĐKXĐ:x\ge0;x\ne1\)
= \(\frac{x+1+\sqrt{x}}{x+1}:\left[\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{x+1+\sqrt{x}}{x+1}:\frac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\frac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
\(=\frac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\frac{x+2}{\sqrt{x}-1}\)
B ) Ta có :
\(Q=P-\sqrt{x}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}-\sqrt{x}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Đế Q nhận giá trị nguyên thì \(1+\frac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\frac{3}{\sqrt{x}-1}\in Z\left(vì1\in Z\right)\)
\(\Leftrightarrow\sqrt{x}-1\inƯ\left(3\right)\)
Ta có bảng sau :
| \(\sqrt{x}-1\) | 3 | -3 | 1 | -1 |
| \(\sqrt{x}\) | 4 | -2 | 2 | 0 |
| \(x\) | 16(t/m) | 4(t/m) | 0(t/m) |
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì \(x\in\left\{16;4;0\right\}\)
\(E=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\) \(\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\sqrt{x}}+\frac{2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\)\(\left[\frac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(E=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(E=\frac{x}{\sqrt{x}-1}\)
b) \(E>1\Leftrightarrow\frac{x}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\frac{x}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{x-2\sqrt{x}+1+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-1\right)^2+\sqrt{x}}{\sqrt{x}-1}>0\)
\(\Rightarrow\sqrt{x}-1>0\) vì tử của phân số luôn \(\ge0\forall x\ge0\)
\(\Rightarrow x>1\)
kết hợp với ĐKXĐ \(x\ge0\Rightarrow x>1\)
vậy \(x>1\) thì \(E>1\)
a) E= \(\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\frac{\sqrt{x}^2-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
= \(\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(a,đkxđ:x>0,x\ne1\)
\(E=\frac{\sqrt{x}.\sqrt{x}}{(\sqrt{x}-1).\sqrt{x}}-\frac{2\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}\)
\(E=\frac{x-2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,ĐểP>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}>0.do\sqrt{x}>0\Rightarrow\)\(\sqrt{x}-1>0\Rightarrow x>1\)
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
a)\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
b)\(E=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=\frac{x-1}{\sqrt{x}}\)
c)Để E>0 thì \(\frac{x-1}{\sqrt{x}}>0\)
Mà \(\sqrt{x}>0\)
\(\Rightarrow x-1>0\) hay \(x>1\)