tìm min P =\(\dfrac{x}{\sqrt{x}-3}\left(x>9\right)\)
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= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Cho x,y,z >0 thỏa x+y+z=\(\sqrt{2021}\)
Tìm Min:
\(P=\sqrt{\left(x+y\right)\left(y+z\right)\left(z+x\right)}.\left(\dfrac{\sqrt{y+z}}{x}+\dfrac{\sqrt{z+x}}{y}+\dfrac{\sqrt{x+y}}{z}\right)\)
Thử nhé
Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)
Thay vo P ta duoc \(P=4.\sqrt{2021}\)
----------------------------------------------------------
\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)
Cauchy-Schwarz:
\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)
\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)
\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)
Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)
\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)
Cho \(P=\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{2\sqrt{x}-x}\right)\)
a) Rút gọn
b) Tìm Min P với \(x>9\)
\(a.P=\left(\dfrac{2+\sqrt{x}}{2-\sqrt{x}}-\dfrac{2-\sqrt{x}}{2+\sqrt{x}}-\dfrac{4x}{x-4}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{2\sqrt{x}-x}\right)=\dfrac{\left(2+\sqrt{x}\right)^2-\left(2-\sqrt{x}\right)^2+4x}{4-x}:\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{8\sqrt{x}+4x}{4-x}.\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}=\dfrac{4\sqrt{x}\left(2+\sqrt{x}\right).\sqrt{x}}{\left(2+\sqrt{x}\right)\left(\sqrt{x}-3\right)}=\dfrac{4x}{\sqrt{x}-3}\) ( x # 4 ; x # 9 ; x > 0 )
\(b.\) \(P=\dfrac{4x}{\sqrt{x}-3}=\dfrac{4\left(x-9\right)+36}{\sqrt{x}-3}=\dfrac{4\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{36}{\sqrt{x}-3}=4\sqrt{x}+12+\dfrac{36}{\sqrt{x}-3}=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Áp dụng BĐT Cauchy cho các số dương , ta có :
\(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}\text{≥}2\sqrt{4\left(\sqrt{x}-3\right).\dfrac{36}{\sqrt{x}-3}}=2.2.6=24\) ⇔ \(4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\) ≥ \(24+24=48\)
⇒ \(P_{MIN}=48."="\text{⇔}x=36\)
cho x,y,z>0 và x+y+z=\(\dfrac{3}{2}\)
tìm Min \(P=\dfrac{\sqrt{x^2+xy+y^2}}{\left(x+y\right)^2+1}+\dfrac{\sqrt{y^2+yz+z^2}}{\left(y+z\right)^2+1}+\dfrac{\sqrt{z^2+zx+x^2}}{\left(z+x\right)^2+1}\)
Đề bài sai, biểu thức này ko có min
Rút gọn \(P=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)rồi tìm x thuộc Z để P thuộc Z.
ĐKXĐ: x>=0; \(x\notin\left\{9;4\right\}\)\(P=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
Để P là số nguyên thì \(3⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;3;-3\right\}\)
=>\(\sqrt{x}\in\left\{3;1;5;-1\right\}\)
=>\(\sqrt{x}\in\left\{3;1;5\right\}\)
=>\(x\in\left\{9;1;25\right\}\)
Kết hợp ĐKXĐ, ta được; \(x\in\left\{1;25\right\}\)
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9; x\neq 4$
\(P=\frac{-3\sqrt{x}+9}{x-9}: \left[\frac{9-x}{(\sqrt{x}-2)(\sqrt{x}+3)}+\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{(\sqrt{x}-2)(\sqrt{x}+3)}-\frac{(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\right]\)
\(=\frac{-3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}:\frac{9-x+x-9-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}\)
\(=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)^2}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{-3}{\sqrt{x}+3}:\frac{-(\sqrt{x}-2)}{\sqrt{x}+3}\\ =\frac{-3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{-(\sqrt{x}-2)}=\frac{3}{\sqrt{x}-2}\)
Với $x\in\mathbb{Z}$, để $P$ nguyên thì $\sqrt{x}-2$ là ước nguyên của 3
$\Rightarrow \sqrt{x}-2\in \left\{1; -1; 3; -3\right\}$
$\Rightarrow \sqrt{x}\in \left\{3; 1; 5; -1\right\}$
$\Rightarrow x\in \left\{9; 1; 25\right\}$
Theo ĐKXĐ suy ra $x=1$ hoặc $x=25$
A =\(\dfrac{x\sqrt[]{x}-3}{x-2\sqrt[]{x}-3}-\dfrac{2\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}+1}+\dfrac{\sqrt[]{x}+3}{3-\sqrt[]{x}}\)
a. rút gọn A
b. Tính A với x = \(14-6\sqrt[]{5}\)
c. tìm min A
a: Ta có: \(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+8}{\sqrt{x}+1}\)
A=\(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
a) Rút gọn A
b) Tính A với x=14-6\(\sqrt{5}\)
c) Tìm Min A
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$
a. \(A=\frac{x\sqrt{x}-3}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{2(\sqrt{x}-3)^2}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)
b.
\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)
\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)
c.
Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$
$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$
Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$
Cho x,y,z > 0 và xyz=8. Tìm Min P = \(\Sigma\dfrac{x^2}{\sqrt{\left(1+x^3\right)+\left(1+y^3\right)}}\)