Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\frac{2\sqrt{a}}{\sqrt{a}+1}>4\Leftrightarrow\frac{2\sqrt{a}}{\sqrt{a}+1}-4>0\Leftrightarrow\frac{2\sqrt{a}-4\sqrt{a}-4}{\sqrt{a}+1}>0\)
\(\Leftrightarrow-2\sqrt{a}-4>0\Leftrightarrow-2\left(\sqrt{a}+2\right)>0\Leftrightarrow\sqrt{a}+2>0\)
\(\Leftrightarrow\sqrt{a}>-2\left(voly\right)\)
#)Giải :
a) \(A=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)\left(\frac{x\sqrt{x}}{\sqrt{x}-1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\frac{x-1}{2\sqrt{x}}\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{2\sqrt{x}}.\frac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\frac{-4}{2\sqrt{x}}=-2\sqrt{x}\)
1)Để căn có nghĩa \(\Leftrightarrow\dfrac{-a}{3}\ge0\Leftrightarrow a\le0\)
Vậy...
2)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2+1}{1-3a}\ge0\\1-3a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-3a>0\left(vìa^2+1>0\right)\\1-3a\ne0\end{matrix}\right.\)
\(\Leftrightarrow1-3a>0\Leftrightarrow3a< 1\Leftrightarrow a< \dfrac{1}{3}\)
Vậy...
3)Để căn có nghĩa
\(\Leftrightarrow a^2-6a+10\ge0\Leftrightarrow\left(a^2-6a+9\right)+1\ge0\Leftrightarrow\left(a-3\right)^2+1\ge0\left(lđ;\forall a\right)\)
Vậy căn luôn có nghĩa với mọi a
4)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-1}{a+2}\ge0\\a+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+2< 0\end{matrix}\right.\end{matrix}\right.\\a+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge1\\a>-2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le1\\a< -2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -2\end{matrix}\right.\)
Vậy...
a) Thay x=25 vào biểu thức \(A=\frac{7}{\sqrt{x}+8}\), ta được:
\(A=\frac{7}{\sqrt{25}+8}=\frac{7}{5+8}=\frac{7}{13}\)
Vậy: khi x=25 thì \(A=\frac{7}{13}\)
b) Ta có: \(B=\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{2\sqrt{x}-24}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x+8\sqrt{x}-3\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+8\right)-3\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}+8}{\sqrt{x}+3}\)
c) Ta có: \(P=A\cdot B\)
\(=\frac{7}{\sqrt{x}+8}\cdot\frac{\sqrt{x}+8}{\sqrt{x}+3}=\frac{7}{\sqrt{x}+3}\)
ĐKXĐ: \(x\ge0\)
Để P có giá trị nguyên thì \(7⋮\sqrt{x}+3\)
\(\Leftrightarrow\sqrt{x}+3\inƯ\left(7\right)\)
\(\Leftrightarrow\sqrt{x}+3\in\left\{1;-7;-1;7\right\}\)
\(\Leftrightarrow\sqrt{x}+3=7\)(vì \(\sqrt{x}+3\ge3\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}=4\)
hay x=16(nhận)
Vậy: Khi x=16 thì P nguyên
d) Ta có: \(\sqrt{x}+3\ge3\forall x\ge0\)
\(\Leftrightarrow\frac{7}{\sqrt{x}+3}\le\frac{7}{3}\forall x\ge0\)
Dấu '=' xảy ra khi x=0
Vậy: Giá trị lớn nhất của biểu thức \(P=A\cdot B\) là \(\frac{7}{3}\) khi x=0
e) Để \(P=\frac{1}{2}\) thì \(\frac{7}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow\sqrt{x}+3=7\cdot2=14\)
\(\Leftrightarrow\sqrt{x}=14-3=11\)
hay x=121(nhận)
Vậy: để \(P=\frac{1}{2}\) thì x=121
a) ĐKXĐ: x\(\ne\) 0;4
Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
= \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)
b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)
<=> \(4=3-\sqrt{x}\)
<=> \(\sqrt{x}=-1\) (vô lí)
Vậy ko tìm được x.
a/A\(=\frac{x+2}{x-\sqrt{x}-2}-\frac{2\sqrt{x}}{\sqrt{x}+1}-\frac{1-\sqrt{x}}{\sqrt{x}-2}\)
\(=\frac{x+2-2\sqrt{x}\left(\sqrt{x}-2\right)-\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2-2x+4\sqrt{x}-1+x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
Thay x=16 vào A ta có: A\(=\frac{3}{2}\)
b/ B= \(1-\frac{\sqrt{x}-3}{\sqrt{x}-2}\)
\(\frac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}=\frac{1}{\sqrt{x}-2}\)
=>C=\(\frac{4\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}:\frac{1}{\sqrt{x}-2}\)=\(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\)
c/Để C thuộc Z thì \(\frac{4\sqrt{x}-1}{\sqrt{x}+1}\) thuộc Z
C\(=\text{}\frac{4\sqrt{x}-1}{\sqrt{x}+1}=\frac{4\sqrt{x}+4}{\sqrt{x}+1}-\frac{5}{\sqrt{x}+1}=4-\frac{5}{\sqrt{x}+1}\)
=> \(5⋮\left(\sqrt{x}+1\right)
\Leftrightarrow\sqrt{x}+1\in\left\{-5;-1;1;5\right\}\)
Nhận xét: \(\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x}+1\in\left\{1;5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;4\right\}
\Leftrightarrow x\in\left\{0;16\right\}\)
Vậy \(x\in\left\{0;16\right\}\) thì C thuộc Z
Chúc bạn học tốt!