tìm gtnn :\(\dfrac{x+3}{\sqrt{x}-1}\)
Giúp mình với!!! Bài này về bất đẳng thức Cauchy ak!!!
1. Cho x > 1 hãy tìm GTNN của:
P=\(\dfrac{x}{\sqrt{x}-1}\)
2. Tìm GTNN của:
B=\(\dfrac{x+15}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\)
\(\left(x\ge0;x\ne1,x\ne9\right)\)
`1. P = x/(sqrt x-1)`
`= (x-1+1)/(sqrtx-1)`
`= ((sqrt x+1)(sqrt x-1))/(sqrt x-1) +1/(sqrt x-1)`
`= sqrt x+1 + 1/(sqrt x-1)`
`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.
ĐTXR `<=> (sqrtx-1)^2 = 1`.
`<=> x =4` hoặc `x = 0 ( ktm)`.
Vậy Min A `= 4 <=> x= 4`.
1) \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{(x-\sqrt{x})+(\sqrt{x}-1)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}+1\)
\(=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)
Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)
Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:
\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)
\(\Rightarrow P\ge2+2=4\)
Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
KL;....
2:
\(B=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}\)
\(=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)
=>\(B>=2\cdot\sqrt{25}-6=4\)
Dấu = xảy ra khi (căn x+3)^2=25
=>căn x+3=5
=>căn x=2
=>x=4
cho P= (\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)+ \(\dfrac{\sqrt{x}}{\sqrt{x-3}}\)-\(\dfrac{3x+3}{x-9}\)) : (\(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}\)-1)
a, Rút gọn P
b, Tìm x để P < \(\dfrac{1}{2}\)
c, Tìm GTNN của P
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
1. Cho \(x,y,z>0\) và \(x^3+y^2+z=2\sqrt{3}+1\). Tìm GTNN của biểu thức \(P=\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\)
2. Cho \(a,b>0\). Tìm GTNN của biểu thức \(P=\dfrac{8}{7a+4b+4\sqrt{ab}}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
1) Áp dụng bđt Cauchy cho 3 số dương ta có
\(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)
\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)
\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)
Cộng (1);(2);(3) theo vế ta được
\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)
\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)
\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)
2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)
Dấu"=" khi a = 4b
nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)
Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)
Đặt \(\sqrt{a+b}=t>0\) ta được
\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)
\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)
Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)
nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)
khi đó a + b = 1
mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)
cho A= \(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
1, rút gọn A, tìm ĐKXĐ
2, tìm x để A< 1
3 Tìm GTNN khi B= (x-9). A
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)
Cho \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
a, Rút gọn P
b, Tìm GTNN của P
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b.
\(P=\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-1+\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}2\sqrt{x}\ge0\\\sqrt{x}+1>0\end{matrix}\right.\) \(\Rightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}\ge0\)
\(\Rightarrow P\ge-1\)
\(P_{min}=-1\) khi \(x=0\)
a) Ta có: \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Tìm GTNN của:
a)\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
Tìm GTLN của:
\(\dfrac{1}{\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}}\)
1:
a: \(A=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
căn x+1>=1
=>2/căn x+1<=2
=>-2/căn x+1>=-2
=>A>=-2+1=-1
Dấu = xảy ra khi x=0
b:
Cho: \(A=\dfrac{3\sqrt{x}}{-x-5\sqrt{x}-1}\)
a) Tìm x biết \(A=\dfrac{2}{3}\)
b) Tìm A biết \(x=7-2\sqrt{6}\)
c) Tìm GTNN của A
b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)
\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)
\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)
cho 2 biểu thức :
\(A=\dfrac{\sqrt{x}+2}{1-\sqrt{x}};B=\left(\dfrac{2\sqrt{x}}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
1, Rút gọn B
2, Đặt P=A.B
Tìm x ∈ Z .Tìm GTNN của P
1: \(B=\dfrac{2\sqrt{x}-x-2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
\(=\dfrac{-x}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-2}\)
Cho biểu thức P = \(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}}{2-\sqrt{x}}\) (với x>0; x\(\ne\)0)
a,Rút gọn biểu thức P và tìm x để P = \(\dfrac{-3}{5}\)
b,Tìm GTNN của biểu thức A=P . \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)
Tìm GTNN A = \(\dfrac{3\sqrt{x}+1}{2\sqrt{x}+2}\) (x ≥ 0)
\(A=\dfrac{3\left(\sqrt{x}+1\right)-2}{2\left(\sqrt{x}+1\right)}=\dfrac{3}{2}-\dfrac{1}{\sqrt{x}+1}\)
Ta có \(\sqrt{x}+1\ge1\Leftrightarrow-\dfrac{1}{\sqrt{x}+1}\ge-1\)
\(\Leftrightarrow A\ge\dfrac{3}{2}-1=\dfrac{1}{2}\)
Dấu \("="\Leftrightarrow x=0\)