Tìm min của biểu thức:\(x+\sqrt{x^2+3}\)
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17 tháng 4 2022
1. 1/x + 2/1-x = (1/x - 1) + (2/1-x - 2) + 3
= 1-x/x + (2-2(1-x))/1-x + 3
= 1-x/x + 2x/1-x + 3 >= 2√2 + 3
Dấu "=" xảy ra khi x =√2 - 1
17 tháng 4 2022
2. a = √z-1, b = √x-2, c = √y-3 (a,b,c >=0)
=> P = √z-1 / z + √x-2 / x + √y-3 / y
= a/a^2+1 + b/b^2+2 + c/c^2+3
a^2+1 >= 2a => a/a^2+1 <= 1/2
b^2+2 >= 2√2 b => b/b^2+2 <= 1/2√2
c^2+3 >= 2√3 c => c/c^2+3 <= 1/2√3
=> P <= 1/2 + 1/2√2 + 1/2√3
Dấu = xảy ra khi a^2 = 1, b^2 = 2, c^2 =3
<=> z-1 = 1, x-2 = 2, y-3 = 3
<=> x=4, y=6, z=2
NT
0
31 tháng 10 2021
\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)
31 tháng 10 2021
a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
AH
Akai Haruma
Giáo viên
17 tháng 11 2021
Lời giải:
a. \(B=\frac{3(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+5}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{3(\sqrt{x}+1)-(\sqrt{x}+5)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}+1}\)
b.
\(P=2AB+\sqrt{x}=2.\frac{\sqrt{x}+1}{\sqrt{x}+2}.\frac{2}{\sqrt{x}+1}+\sqrt{x}=\frac{4}{\sqrt{x}+2}+\sqrt{x}\)
Áp dụng BĐT Cô-si:
$P=\frac{4}{\sqrt{x}+2}+(\sqrt{x}+2)-2\geq 2\sqrt{4}-2=2$
Vậy $P_{\min}=2$ khi $\sqrt{x}+2=2\Leftrightarrow x=0$
7 tháng 8 2016
diều kiện x >= 0
P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)
= \(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)
=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)
P=8/9
<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)
<=> \(2x-5\sqrt{x}+2=0\)
<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)
vậy x=4 hoặc x=1/4 thì p=8/9
7 tháng 8 2016
a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)
\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P=8/9
\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)
\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)
\(\Leftrightarrow24x-60\sqrt{x}+24=0\)
\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)
BA
0
24 tháng 6 2021
a) đk: x\(\ge0\);
P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)
= \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Để P = \(\dfrac{8}{9}\)
<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)
<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)
<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)
<=> \(-2x+5\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)
c)
Đặt \(\sqrt{x}=a\) (\(a\ge0\))
P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)
Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)
Dấu "=" <=> a = -1 (loại)
=> Không tìm được Min của P
Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)
<=> \(P\le\dfrac{4}{3}\)
Dấu "=" <=> a = 1 <=> x = 1 (tm)