min P=\(\dfrac{x+8}{\sqrt{x}+1}\)
Bài 1:Cho x\(\ge0\).Tìm giá trị nhỏ nhất hoặc giá trị lớn nhất của các biểu thức sau:
1)A=3x+2\(\sqrt{x}\)+1min
2)A=x+3\(\sqrt{x}\)-3min
3)A=-2x-3\(\sqrt{x}\)+2max
4)A=-4x-5\(\sqrt{x}\)-3max
5)A=x-2\(\sqrt{x}\)+2min
6)A=x-4\(\sqrt{x}\)-5min
7)A=-x+6\(\sqrt{x}\)+5max
8)A=-x+8\(\sqrt{x}\)-10max
9)A=\(\dfrac{2}{\sqrt{x}+1}\)max
10)A=\(\dfrac{4}{\sqrt{x}+2}\)max
11)A=\(\dfrac{-3}{\sqrt{x}+3}\)min
12)A=\(\dfrac{-5}{\sqrt{x}+4}\)min
13)A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)max
14)A=\(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\)max
15)A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)min
16)A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+4}\)min
17)A=\(\dfrac{x+3}{\sqrt{x}+1}\)min
18)A=\(\dfrac{x+5}{\sqrt{x}+2}\)min
19)A=\(\dfrac{x+12}{\sqrt{x}+2}\)min
20)A=\(\dfrac{x+7}{\sqrt{x}+3}\)min
21)A=\(\dfrac{x+9}{\sqrt{x}+4}\)min
22)A=\(\dfrac{x+24}{\sqrt{x}+5}\)min
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1
cho biểu thức P =\(\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\times\dfrac{4\sqrt{x}}{3}\) với x ≥ 0
a, Rút gọn P,
b, Tìm x để P=\(\dfrac{8}{9}\),
c, Tìm Max và Min của P
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)
b) Ta có: \(P=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)
\(=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
Ta có: \(P=\dfrac{8}{9}\)
nên \(36\sqrt{x}=27\left(x-\sqrt{x}+1\right)\)
\(\Leftrightarrow27x-27\sqrt{x}+27-36\sqrt{x}=0\)
\(\Leftrightarrow27x-63\sqrt{x}+27=0\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
M=\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)
tính Min M
ĐKXĐ: \(x>0;x\ne1\)
\(M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x}\)
\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(x-1\right)}{\sqrt{x}}.\dfrac{x}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(M_{min}=-\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)
cho A=\(\dfrac{x+1}{x\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
a, Rút gọn A
b, tính A khi x=\(33-8\sqrt{2}\)
c , CMR: A< \(\dfrac{1}{3}\)
d, tìm Min B = \(\dfrac{1}{A}\)
Tìm Min P= \(\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\)
\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{x-1}\right):\dfrac{1}{\sqrt{x}-1}\)
ĐKXĐ:
\(x\ne1\)
\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{\left(\sqrt{x}\right)^2-1}\right):\dfrac{1}{\sqrt{x}-1}\)
\(P=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{1}{\sqrt{x}-1}\)\(P=\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{1}{\sqrt{x}-1}\)\(P=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)-x-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(P=\dfrac{x+3\sqrt{x}+2-x-4}{\sqrt{x}+1}\)
\(P=\dfrac{3\sqrt{x}-4}{\sqrt{x}+1}\)
\(=\dfrac{3\sqrt{x}+3-5}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)-5}{\sqrt{x}+1}=3-\dfrac{5}{\sqrt{x}+1}\)
Với mọi giá trị của x ta có:
\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{5}{\sqrt{x}+1}\le5\)
\(\Rightarrow P\ge3-5=-2\)
Vậy \(Min_P=-2\)
Để P = -2 thì \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)
Cho x,y,z > 0 và xyz=8. Tìm Min P = \(\Sigma\dfrac{x^2}{\sqrt{\left(1+x^3\right)+\left(1+y^3\right)}}\)
cho x,y>0 và x+y=1 tìm min p=\(\dfrac{x}{\sqrt{1-x}}\)+\(\dfrac{y}{\sqrt{1-y}}\)
Lời giải:
Do $x+y=1$ nên:
$P=\frac{x}{\sqrt{x+y-x}}+\frac{y}{\sqrt{x+y-y}}=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}$
$=\frac{x^2}{x\sqrt{y}}+\frac{y^2}{y\sqrt{x}}$
$\geq \frac{(x+y)^2}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{x\sqrt{y}+y\sqrt{x}}$ (áp dụng BĐT Cauchy-Schwarz)
Áp dụng BĐT Bunhiacopxky:
$(x\sqrt{y}+y\sqrt{x})^2\leq (x+y)(xy+xy)=2xy(x+y)\leq \frac{(x+y)^2}{2}(x+y)=\frac{1}{2}$
$\Rightarrow x\sqrt{y}+y\sqrt{x}\leq \frac{\sqrt{2}}{2}$
$\Rightarrow P\geq \frac{1}{x\sqrt{y}+y\sqrt{x}}\geq \frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}$
Vậy $P_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=y=\frac{1}{2}$.
Cho P=\(\left(\dfrac{2x\sqrt{x}+x}{x\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{x-1}\right).\dfrac{x+\sqrt{x}}{x-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
a.Rút gọn P
b.Tìm MinP