K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
27 tháng 6 2021
`D=(sqrt{3}.sqrt{5-2sqrt6})/(sqrt3-sqrt2)-1/(2-sqrt3)`
`=(sqrt3*sqrt{3-2sqrt{3}.sqrt2+2})/(sqrt3-sqrt2)-(2+sqrt3)/(4-3)`
`=(sqrt3.sqrt{(sqrt3-sqrt2)^2})/(sqrt3-sqrt2)-2-sqrt3`
`=sqrt3-2-sqrt3=-2`
21 tháng 9 2021
\(b,B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\left(x\ge0;x\ne4;x\ne9\right)\\ B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
\(c,B< A\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< \dfrac{\sqrt{x}+1}{\sqrt{x}-2}\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 0\\ \Leftrightarrow\dfrac{-5}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\left(-5< 0\right)\\ \Leftrightarrow x>4\\ d,P=\dfrac{B}{A}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}:\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}-4}{\sqrt{x}+1}=1-\dfrac{5}{\sqrt{x}+1}\in Z\\ \Leftrightarrow5⋮\sqrt{x}+1\Leftrightarrow\sqrt{x}+1\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{-6;-2;0;4\right\}\\ \Leftrightarrow x\in\left\{0;16\right\}\left(\sqrt{x}\ge0\right)\)
\(e,P=1-\dfrac{5}{\sqrt{x}+1}\)
Ta có \(\sqrt{x}+1\ge1,\forall x\Leftrightarrow\dfrac{5}{\sqrt{x}+1}\ge5\Leftrightarrow1-\dfrac{5}{\sqrt{x}+1}\le-4\)
\(P_{max}=-4\Leftrightarrow x=0\)
16 tháng 7 2021
Bài 2:
d) Ta có: \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
e) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
2 tháng 10 2021
a) \(P=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{9}+5}{\sqrt{9}-2}=\dfrac{3+5}{3-2}=8\)
b) \(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{5\sqrt{x}-2}{4-x}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c) \(M=\dfrac{Q}{P}=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{\sqrt{x}-2}{\sqrt{x}+5}=\dfrac{\sqrt{x}}{\sqrt{x}+5}< \dfrac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}< 3\sqrt{x}+15\Leftrightarrow\sqrt{x}>-15\left(đúng\forall x\ge0,x\ne4\right)\)
d) \(M=\dfrac{\sqrt{x}}{\sqrt{x}+5}=1-\dfrac{5}{\sqrt{x}+5}\in Z\)
\(\Rightarrow\sqrt{x}+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(x\ge0,x\ne4\)
\(\Rightarrow x\in\left\{0\right\}\)
20 tháng 10 2021
Bài 6:
a: Ta có: \(E=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
B2
0
28 tháng 6 2021
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}`
1.
d, ĐK: \(x\ge-5\)
\(x-2-4\sqrt{x+5}=-10\)
\(\Leftrightarrow x+5-4\sqrt{x+5}+3=0\)
\(\Leftrightarrow\left(\sqrt{x+5}-1\right)\left(\sqrt{x+5}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1\\\sqrt{x+5}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=1\\x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
\(\Leftrightarrow x=\pm4\left(tm\right)\)
2.
ĐK: \(x\in R\)
\(\sqrt{x^2+2x+1}+\sqrt{x^2-4x+4}=3\)
\(\Leftrightarrow\sqrt{\left(x+1\right)^2}+\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left|x+1\right|+\left|x-2\right|=3\)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\).
\(\left|x+1\right|+\left|x-2\right|=\left|x+1\right|+\left|2-x\right|\ge\left|x+1+2-x\right|=3\)
Đẳng thức xảy ra khi:
\(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Leftrightarrow-1\le x\le2\)