Tìm GTNN :
A=\(\frac{5x-4\sqrt{x}+1}{x}\)
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BT
10 tháng 8 2017
\(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a/ \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt[]{x-3}\right)}\right]:\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3}{\sqrt[]{x-3}}\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right]:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
=> \(R=\left[\frac{2\sqrt{x}+\sqrt{x}-3}{\sqrt{x}-3}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=> \(R=\frac{3\sqrt{x}-3}{\sqrt{x}-3}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
b/ Để R<-1 => \(\frac{3\left(\sqrt{x}-1\right)}{\sqrt{x}+1}< -1\)
<=> \(3\sqrt{x}-3< -\sqrt{x}-1\)
<=> \(4\sqrt{x}< 2\)=> \(\sqrt{x}< \frac{1}{2}\) => \(-\frac{1}{4}< x< \frac{1}{4}\)
10 tháng 8 2017
Chỗ => R = \(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}+1\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\) là sao vậy ạ?
0
6
8 tháng 1 2019
a,\(\frac{x}{\sqrt{x}+1}=\frac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}\)
\(=\left(\sqrt{x}-1\right)+\frac{1}{\sqrt{x}-1}+2\ge2.\sqrt{\left(\sqrt{x}-1\right).\frac{1}{\sqrt{x}-1}+2}\ge4\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-1=\frac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}-1=1\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\left(t/m\right)\)
Dmin = 4 <=> x=4
b,\(\frac{\sqrt{x-9}}{5x}\)
\(\sqrt{x-9}=\sqrt{\frac{\left(x-9\right).9}{9}}=\frac{1}{3}.\sqrt{\left(x-9\right).9}\le\frac{1}{3}.\frac{x-9+9}{2}=\frac{x}{2}\)
\(\Rightarrow D\le\frac{x}{\frac{6}{5x}}=\frac{x}{30x}=\frac{1}{30}\)
Dấu "=" xảy ra \(\Leftrightarrow x-9=9\Leftrightarrow x=18\)
Dmax=\(\frac{1}{30}\Leftrightarrow x=18\)
P/s : ko chắc lắm
8 tháng 1 2019
\(a)\)\(P=\frac{x}{\sqrt{x}+1}=\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\frac{2\sqrt{x}+2}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}-\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\frac{1}{\sqrt{x}+1}\)
\(P=\sqrt{x}+1+\frac{1}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{1}{\sqrt{x}+1}}-2=2-2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}+1=\frac{1}{\sqrt{x}+1}\)\(\Leftrightarrow\)\(x=0\)
...
PT
0
6 tháng 6 2018
Điều kiện có 2 nghiệm phân biệt tự làm nha
Theo vi-et ta có:
\(\hept{\begin{cases}x_1+x_2=5\\x_1.x_2=m-2\end{cases}}\)
\(2\left(\frac{1}{\sqrt{x_1}}+\frac{1}{\sqrt{x_2}}\right)=3\)
\(\Leftrightarrow4\left(\frac{1}{x_1}+\frac{1}{x_2}+\frac{2}{\sqrt{x_1.x_2}}\right)=9\)
\(\Leftrightarrow4\left(\frac{5}{m-2}+\frac{2}{\sqrt{m-2}}\right)=9\)
Làm nốt nhé
6 tháng 6 2018
Câu 1:
M=\(\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(4x^2-4x+1\right)+2014\)
=\(\left(\left(x+y\right)^2+2\left(x+y\right)+1\right)+\left(2x-1\right)^2+2014\)
=\(\left(x+y+1\right)^2+\left(2x-1\right)^2+2014\ge2014\)
\(\Rightarrow M\ge2014\Leftrightarrow minM=2014\)
\(\Leftrightarrow\hept{\begin{cases}x+y+1=0\\2x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0,5\\y=1,5\end{cases}}\)
12 tháng 1 2021
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{3\sqrt{x}+1}{1-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) Với x = 4 thỏa mãn ĐKXĐ
\(A=\frac{2\sqrt{4}-1}{\sqrt{4}+1}=\frac{4-1}{2+1}=\frac{3}{3}=1\)
c) Chưa nghĩ ra :<
AN
10 tháng 10 2019
\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow x>0\)
Ta có:
\(\frac{5x-4\sqrt{x}+1}{x}=\frac{x+4x-4\sqrt{x}+1}{x}=1+\frac{\left(2\sqrt{x}-1\right)^2}{x}\ge1\)(vì \(x>0\))
Dấu "=" xảy ra khi \(2\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\left(tm\right)\)
Vậy \(Min_{bt}=1\) khi \(x=\frac{1}{4}\)
2 tháng 10 2015
A = \(\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)
= \(\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
= \(\sqrt{x-4}+2+l\sqrt{x-4}-2l\)
(+) với \(l\sqrt{x-4}-2l=\sqrt{x-4}-2\) khi \(x\ge8\)
=> A = \(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
(+) \(l\sqrt{x-4}-2l=2-\sqrt{x-4}\) khi \(4\le x\le8\)
=> A = \(\sqrt{x-4}+2+2-\sqrt{x-4}=4\)
2 tháng 10 2015
1) Áp dụng bất đẳng thức Cô - si với 4 số \(\frac{5x}{3};\frac{5x}{3};\frac{5x}{3};\frac{1}{x^3}\) dương ta có:
\(B=\frac{5x}{3}+\frac{5x}{3}+\frac{5x}{3}+\frac{1}{x^3}\ge4\sqrt[4]{\frac{5x}{3}.\frac{5x}{3}.\frac{5x}{3}.\frac{1}{x^3}}=4\sqrt[4]{\frac{125}{27}}\)
=> B nhỏ nhất bằng \(4\sqrt[4]{\frac{125}{27}}\) khi \(\frac{5x}{3}=\frac{1}{x^3}\) => x4 = 3/5 => x = \(\sqrt[4]{\frac{3}{5}}\)
2) ĐK : x > 4
\(A=\sqrt{\left(x-4\right)+2\sqrt{x-4}.2+4}+\sqrt{\left(x-4\right)-2\sqrt{x-4}.2+4}\)
\(A=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}\)
\(A=\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\)
+) Nếu \(\sqrt{x-4}\ge2\) => x - 4 > 4 => x > 8 thì \(A=\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
+) Nếu \(\sqrt{x-4}
ĐKXĐ : x > 0
Ta có : \(A=\frac{5x-4\sqrt{x}+1}{x}=5-\frac{4}{\sqrt{x}}+\frac{1}{x}\)
Đặt : \(t=\frac{1}{\sqrt{x}}>0\) Khi đó : \(A=5-4t+t^2=\left(t^2-4t+4\right)+1=\left(t-2\right)^2+1\ge1\)
" = " <=> t = 2 \(\Leftrightarrow\frac{1}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)
Vậy ..
ĐKXĐ : x > 0
\(A=\frac{5x-4\sqrt{x}+1}{x}=1+\frac{4x-4\sqrt{x}+1}{x}=1+\frac{\left(2\sqrt{x}-1\right)^2}{x}\ge1\)
=> Min A = 1
Dấu "=" xảy ra <=> \(2\sqrt{x}-1=0\)
<=> \(\sqrt{x}=\frac{1}{2}\)
<=> x = 1/4
Vậy Min A = 1 <=> x = 1/4