Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: ĐKXĐ: x>=0; x<>1
\(P=\dfrac{-3+\sqrt{x}-1}{x-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-4}{\sqrt{x}-1}\)
b: Để P=5/4 thì \(\dfrac{\sqrt{x}-4}{\sqrt{x}-1}=\dfrac{5}{4}\)
=>\(5\sqrt{x}-5=4\sqrt{x}-16\)
=>căn x=-11(loại)
a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
b. M =\(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{-1}{\sqrt{x}+1}\)
c. \(M=\frac{-1}{\sqrt{x}+1}\ge-1\)
Vậy Min M =-1 khi x=0
\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=2\)
=> Với mọi \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)thì P = 2
Đề sai à --
\(a,ĐK:x>0;x\ne1;x\ne4\\ b,P=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{x-1-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\\ P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{3}=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\)
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 \(x\in\left\{16;4;0\right\}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)
c: Ta có: P=AB
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}+1}\)