LH
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HT
15 tháng 6 2019
a/ ĐKXĐ:...
\(N=\left(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right).\frac{4\sqrt{x}}{3}\)
\(N=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\frac{4\sqrt{x}}{3}\)
\(N=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b/ N=\(N=\frac{8}{9}\Rightarrow\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
\(\Leftrightarrow36\sqrt{x}=24x-24\sqrt{x}+24\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
Đặt \(\sqrt{x}=a\ge0\Rightarrow x=a^2\)
\(\Rightarrow24a^2-60a+24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Y
15 tháng 6 2019
\(a.N=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\left(\frac{x+2}{\sqrt{x^3}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\left(\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\frac{4\sqrt{x}}{3}\\ =\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
\(b.N=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\Leftrightarrow4\sqrt{x}=\frac{8}{3}\left(x-\sqrt{x}+1\right)\\ \Leftrightarrow3\sqrt{x}=2\left(x-\sqrt{x}+1\right)\\ \Leftrightarrow2x-5\sqrt{x}+2=0\\ \left[{}\begin{matrix}x=4\\x=\frac{1}{4}\end{matrix}\right.\)
Bước cuối bạn tự làm nha (do mk bận)
S
13 tháng 10 2019
\(\text{Áp dụng BĐT:}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)
\(\frac{1}{3x+3y+2z}=\frac{1}{\left(x+y\right)+\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(\text{tương tự với các BĐT còn lại }\)
\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+3z+2y}+\frac{1}{3y+3z+2x}\le\frac{1}{16}.\left(\frac{4}{x+z}+\frac{4}{x+y}+\frac{4}{y+z}\right)=\frac{1}{16}.24=\frac{3}{2}\left(đpcm\right)\)
Y
20 tháng 6 2019
a)
\(N=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\\ =\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\\ =x-\sqrt{x}+1\)
b)
\(N=x-\sqrt{x}+1=x-2\cdot\sqrt{x}\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy Min N = \(\frac{3}{4}\)khi x=\(\frac{1}{4}\)
Câu c) mk ko bt, sorry nha :<
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2019
\(M=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{2\sqrt{x}}{\left(x-\frac{1}{2}\right)^2+\frac{3}{4}}>0\) \(\forall x>0\)
\(M-2=\frac{2\sqrt{x}}{x-\sqrt{x}+1}=\frac{-2\left(x-2\sqrt{x}+1\right)}{x-\sqrt{x}+1}=\frac{-2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}< 0\) \(\forall x\ne1;x>0\)
\(\Rightarrow0< M< 2\Rightarrow\) để M nguyên thì \(M=1\)
\(\Rightarrow\frac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow x-3\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3\pm\sqrt{5}}{2}\) \(\Rightarrow x=\frac{7\pm3\sqrt{5}}{2}\)
DH
3 tháng 2 2020
\(M\left(x+y+z\right)=\left(z^2+y^2+z^2\right)+2+\frac{\left(x^2+1\right)\left(y+z\right)}{x}+\frac{\left(y^2+1\right)\left(z+x\right)}{y}+\frac{\left(z^2+1\right)\left(x+y\right)}{z}\)
\(=5+\frac{\left(x^2+1\right)\left(y+z\right)}{x}+\frac{\left(y^2+1\right)\left(z+x\right)}{y}+\frac{\left(z^2+1\right)\left(x+y\right)}{z}\)
\(\ge5+2\left(y+z\right)+2\left(z+x\right)+2\left(x+y\right)=5+4\left(x+y+z\right)\) ( Sử dụng BĐT Cô-si cho 2 số dương ý)
\(\Rightarrow M\ge\frac{5}{x+y+z}+4\)
Mặt khác: \(\left(x+y+z\right)^2\le\left(x^2+y^2+z^2\right)\left(1^2+1^2+1^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
Do đó: \(M\ge\frac{5}{3}+4=\frac{17}{3}\)
\(M=\frac{17}{3}\Leftrightarrow x=y=z=1\)
\(\Rightarrow Min_A=\frac{17}{3}\)