![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐIều kiện:`x^2-7x+8>=0`
`<=>x^2-2*x*7/2+49/4-17/4>=0`
`<=>(x-7/2)^2-17/4>=0`
`<=>(x-7/2)^2>=17/4`
`<=>|x-7/2|>=sqrt{17}/2`
`<=>` \(\left[ \begin{array}{l}x \ge \dfrac{7+\sqrt{17}}{2}\\x \le \dfrac{-\sqrt{17}+7}{2}\end{array} \right.\)
`pt<=>x^2-7x+sqrt{x^2-7x+8}-12=0`
`<=>x^2-7x+8+sqrt{x^2-7x+8}-20=0`
Đặt `a=sqrt{x^2-7x+8}(a>=0)`
`pt<=>a^2+a-20=0`
`<=>a=4(tm),a=-5(l)`
`<=>x^2-7x+8=16`
`<=>x^2-7x-8=0`
`a-b+c=0`
`=>x_1=-1(tm),x_2=8(tm)`
Vậy `S={-1,8}`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\dfrac{1}{x^2-2x+1}}=\sqrt{\dfrac{1}{\left(x-1\right)^2}}\)
ĐKXĐ: \(\left(x-1\right)^2\ne0\Leftrightarrow x\ne1\)
Vậy \(x\in R,x\ne1\) thì căn thức xác định
![](https://rs.olm.vn/images/avt/0.png?1311)
Biểu thức này không có GTLN vì nếu cho x > 1 và \(x\rightarrow1\Rightarrow\dfrac{1}{\sqrt{x}-1}\rightarrow\infty\).
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
12:
a: ĐKXĐ: -2x+3>=0
=>x<=3/2
b: ĐKXĐ: x^2<>0
=>x<>0
c: ĐKXĐ: x+3>0
=>x>-3
d:ĐKXĐ: -5/x^2+6>=0
=>x^2+6<0
=>x thuộc rỗng
4.
a) Ta có: \(\sqrt{x}=3\)
\(\Leftrightarrow x=3^2=9\)
Vậy \(x=9\)
b) Ta có: \(\sqrt{x}=\sqrt{5}\)
\(\Leftrightarrow x=\sqrt{5}^2=5\)
Vậy \(x=5\)
c) Ta có: \(\sqrt{x}=0\)
\(\Leftrightarrow x=0^2=0\)
Vậy \(x=0\)
d) Ta có: \(\sqrt{x}=-2\)
\(\Leftrightarrow x=-2^2=4\)
Vậy \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(P=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b: \(P-1=\dfrac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\dfrac{-1}{\sqrt{a}}< 0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(19,ĐKXĐ:\dfrac{-2\sqrt{6}+\sqrt{23}}{-x+5}\ge0;x\ne5\\ \Leftrightarrow5-x< 0\left(-2\sqrt{6}+\sqrt{23}< 0\right)\\ \Leftrightarrow x>5\)
\(21,ĐKXĐ:x\ne7;\dfrac{2\sqrt{15}-\sqrt{59}}{x-7}\ge0\\ \Leftrightarrow x-7>0\left(2\sqrt{15}-\sqrt{59}>0\right)\\ \Leftrightarrow x>7\)
\(23,ĐKXĐ:49x^2-24x+4\ge0\Leftrightarrow\left(49x^2-14\cdot\dfrac{12}{7}x+\dfrac{144}{49}\right)+\dfrac{52}{49}\ge0\\ \Leftrightarrow\left(7x-\dfrac{12}{7}\right)^2+\dfrac{52}{49}\ge0\\ \Leftrightarrow x\in R\)