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NM
Nguyễn Minh Quang
Giáo viên
20 tháng 3 2021
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(\Leftrightarrow P=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)
b. ta có \(x=\frac{8-4\sqrt{3}}{2-\sqrt{3}}=4\)
vậy \(P=\frac{4}{\sqrt{4}-1}=4\)
c.\(P=\frac{x}{\sqrt{x}-1}=\sqrt{x}-1+\frac{1}{\sqrt{x}-1}+2\ge2+2=4\)
vậy \(\sqrt{P}\ge2\)
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 \(x=\frac{1}{4}\)
24 tháng 7 2019
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
NM
Nguyễn Minh Quang
Giáo viên
2 tháng 9 2021
để \(y=\left(\sqrt{3}-\sqrt{5}\right)x+\sqrt{5}+\sqrt{3}=1\)
thì \(\left(\sqrt{3}-\sqrt{5}\right)x=1-\sqrt{5}-\sqrt{3}\)
\(\Leftrightarrow x=\frac{1-\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}\)
b.\(f^2\left(x\right)=\left[\left(\sqrt{3}-\sqrt{5}\right)x+\sqrt{5}+\sqrt{3}\right]^2=8+2\sqrt{15}=\left(\sqrt{5}+\sqrt{3}\right)^2\)
\(\Leftrightarrow\left[\left(\sqrt{3}-\sqrt{5}\right)x+2\sqrt{5}+2\sqrt{3}\right]\left(\sqrt{3}-\sqrt{5}\right)x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{2\left(\sqrt{3}+\sqrt{5}\right)x}{\left(\sqrt{3}-\sqrt{5}\right)x}\end{cases}}\)
22 tháng 6 2016
bài 1
a) ĐKXĐ : bạn tự tìm nhé
b) ta có A=\(\sqrt{x^2-1+2\sqrt{x^2-1}+1}-\sqrt{x^2-1-2\sqrt{x^2-1}+1}\)
=\(\sqrt{\left(\sqrt{x^2-1}+1\right)^2}+\sqrt{\left(\sqrt{x^2-1}-1\right)^2}\)
=\(\left|\sqrt{x^2-1}+1\right|+\left|\sqrt{x^2-1}-1\right|\)
=\(\sqrt{x^2-1}+1+\sqrt{x^2-1}-1\)( vì \(\left|x\right|\ge\sqrt{2}\))
=\(2\sqrt{x^2-1}\)
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
\(A=x-\sqrt{x}\)
\(v\text{ì:}x\ge\sqrt{x}\Rightarrow x-\sqrt{x}\ge0\)
\(d\text{ấu}``=''\Leftrightarrow x=0\)
Vậy minA=0 tại x=0