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3 tháng 3 2023
a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
b: \(\sqrt{x}+3>=3\)
=>A<=1
Dấu = xảy ra khi x=0
c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)
Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{1;25\right\}\)
16 tháng 5 2019
2. \(P=x^2-x\sqrt{3}+1=\left(x^2-x\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=\left(x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)
Dấu '=' xảy ra khi \(x=\frac{\sqrt{3}}{2}\)
Vây \(P_{min}=\frac{1}{4}\)khi \(x=\frac{\sqrt{3}}{2}\)
3. \(Y=\frac{x}{\left(x+2011\right)^2}\le\frac{x}{4x.2011}=\frac{1}{8044}\)
Dấu '=' xảy ra khi \(x=2011\)
Vây \(Y_{max}=\frac{1}{8044}\)khi \(x=2011\)
4. \(Q=\frac{1}{x-\sqrt{x}+2}=\frac{1}{\left(x-\sqrt{x}+\frac{1}{4}\right)+\frac{7}{4}}=\frac{1}{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}\le\frac{4}{7}\)
Dấu '=' xảy ra khi \(x=\frac{1}{4}\)
Vậy \(Q_{max}=\frac{4}{7}\)khi \(x=\frac{1}{4}\)
D
0
30 tháng 5 2023
a: \(P=\dfrac{\sqrt{x}+1-2\sqrt{x}+4+2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{\sqrt{x}+1}\)
b: căn x+1>=1
=>P<=1
Dấu = xảy ra khi x=0
29 tháng 7 2021
I) Đk: x > 0 và x \(\ne\)9
\(D=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(D=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)
\(D=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
=> \(\frac{1}{D}=\frac{\sqrt{x}+3}{\sqrt{x}+1}=\frac{\sqrt{x}+1+2}{\sqrt{x}+1}=1+\frac{2}{\sqrt{x}+1}\)
Để 1/D nguyên <=> \(\frac{2}{\sqrt{x}+1}\in Z\)
<=> \(2⋮\left(\sqrt{x}+1\right)\) <=> \(\sqrt{x}+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
Do \(x>0\) => \(\sqrt{x}+1>1\) => \(\sqrt{x}+1=2\)
<=> \(\sqrt{x}=1\) <=> x = 1 (tm)
29 tháng 7 2021
\(E=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)
\(E=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
b) Với x\(\ge\)0; ta có:
\(E=\frac{8}{9}\) <=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)
<=> \(3\sqrt{x}=2x-2\sqrt{x}+2\)
<=> \(2x-4\sqrt{x}-\sqrt{x}+2=0\)
<=> \(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
e) Ta có: \(E=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\ge0\forall x\in R\) (vì \(x-\sqrt{x}+1=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\))
Dấu "=" xảy ra<=> x = 0
Vậy MinE = 0 <=> x = 0
Lại có: \(\frac{1}{E}=\frac{3\left(x-\sqrt{x}+1\right)}{4\sqrt{x}}=\frac{3}{4}\left(\sqrt{x}-1+\frac{1}{\sqrt{x}}\right)\ge\frac{3}{4}\left(2\sqrt{\sqrt{x}\cdot\frac{1}{\sqrt{x}}}-1\right)\)(bđt cosi)
=> \(\frac{1}{E}\ge\frac{3}{2}.\left(2-1\right)=\frac{3}{2}\)=> \(E\le\frac{2}{3}\)
Dấu "=" xảy ra<=> \(\sqrt{x}=\frac{1}{\sqrt{x}}\) <=> x = 1
Vậy MaxE = 2/3 <=> x = 1
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2020
\(A=\frac{1.\sqrt{x-1}}{x}+\frac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}y}+\frac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}z}\)
\(A\le\frac{1}{2}\left(\frac{1+x-1}{x}+\frac{2+y-2}{\sqrt{2}y}+\frac{3+z-3}{\sqrt{3}z}\right)=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right)\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;4;6\right)\)
3 tháng 2 2022
a: \(P=\dfrac{x+\sqrt{x}+1+11\sqrt{x}-11+34}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+12\sqrt{x}+24}{\sqrt{x}+2}\)
b: Thay \(x=3-2\sqrt{2}\) vào P, ta được:
\(P=\dfrac{3-2\sqrt{2}+12\left(\sqrt{2}-1\right)+24}{\sqrt{2}-1+2}\)
\(=\dfrac{27-2\sqrt{2}+12\sqrt{2}-12}{\sqrt{2}+1}=5+5\sqrt{2}\)
Ta có: C = \(\frac{\sqrt{x}-1}{x-3\sqrt{x}+11}=\frac{\left(x-3\sqrt{x}+11\right)-\left(x-8\sqrt{x}+16\right)}{5\left(x-3\sqrt{x}+11\right)}=\frac{1}{5}-\frac{\left(\sqrt{x}-4\right)^2}{5\left(x-3\sqrt{x}+11\right)}\)
Do: \(\frac{\left(\sqrt{x}-4\right)^2}{x-3\sqrt{x}+11}\ge0\)(\(x-3\sqrt{x}+11=\left(\sqrt{x}-\frac{3}{2}\right)^2+\frac{35}{4}>0\))
=> \(-\frac{\left(\sqrt{x}-4\right)^2}{x-3\sqrt{x}+11}\le0\) => \(\frac{1}{5}-\frac{\left(\sqrt{x}-4\right)^2}{x-3\sqrt{x}+11}\le\frac{1}{5}\) => C \(\le\)1/5
Dấu "=" xảy ra <=> \(\sqrt{x}-4=0\)<=> x = 16
Vậy Max C = 1/5 <=> x = 16