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\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)ĐK x>=0 x khác -1
=\(\frac{\sqrt{x}+1}{x+1}.\frac{x+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b/ x =\(\frac{2+\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{4}=\frac{3+2\sqrt{3}+1}{4}=\frac{\left(\sqrt{3}+1\right)^2}{4}\)
\(\Rightarrow\sqrt{x}=\frac{\sqrt{3}+1}{2}\)
Em thay vào tính nhé!
c) với x>1
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-1}.\sqrt{x}=\frac{x+\sqrt{x}}{\sqrt{x}-1}=\sqrt{x}+2+\frac{2}{\sqrt{x}-1}=\sqrt{x}-1+\frac{2}{\sqrt{x}-1}+3\)
Áp dụng bất đẳng thức Cosi
A\(\ge2\sqrt{2}+3\)
Xét dấu bằng xảy ra ....
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
điều kiện xác định : \(x>0;x\ne4\)
a) ta có : \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(\Leftrightarrow P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\right)\) \(\Leftrightarrow P=\left(\dfrac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\left(\sqrt{x}-2\right)^2-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\) \(\Leftrightarrow P=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{4}\right)\) \(\Leftrightarrow P=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}\)b) để \(P>0\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}>0\) \(\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-3>0\\\sqrt{x}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-3< 0\\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>9\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 9\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 1\end{matrix}\right.\)
kết hợp với điều kiện xác định ta có : \(0< x< 1\) hoặc \(x>9\)
c) ta có : \(\sqrt{P}=\sqrt{\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}}\ge0\forall x\)
dấu "=" xảy ra khi \(\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
vậy ....................................................................................................
d) ta có : \(m\left(\sqrt{x}-3\right)P=12m\sqrt{x}-4\)
\(\Leftrightarrow m\left(\sqrt{x}-3\right)\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{4}=12m\sqrt{x}-4\)
\(\Leftrightarrow m\left(x-6\sqrt{x}+9\right)\left(\sqrt{x}-1\right)=48m\sqrt{x}-4\)
nhân tung ra giải bình thường ............(mk nghỉ có vấn đề ở câu d này nha )
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
a) Thay x=4 zô là đc . ra kết quả \(\frac{7}{6}\)là dúng
b) \(B=\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=\frac{3x+3\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\)
\(=>P=A.B=\frac{3\sqrt{x}+1}{x+\sqrt{x}}.\frac{3\left(x+\sqrt{x}\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}=\frac{3}{3\sqrt{x}-1}\)
c) xét \(\frac{1}{P}=\frac{3\sqrt{x}-1}{3}\)
do \(\sqrt{x}\ge0=>3\sqrt{x}-1\ge-1\)\(=>\frac{3\sqrt{x}-1}{3}\ge-\frac{1}{3}\)
\(=>\frac{1}{P}\ge-\frac{1}{3}\)
dấu = xảy ra khi x=0
zậy ..
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4\sqrt{x}-3}{2\sqrt{x}-x}\right):\)\(\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}-4}{\sqrt{x}-2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)\(:\left(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{x-4-x+4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{4\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}-3}{4}\)
\(b,\)Để \(P>0\Rightarrow\frac{\sqrt{x}-3}{4}>0\)
Mà \(4>0\Rightarrow\sqrt{x}-3>0\Rightarrow\sqrt{x}>3\Rightarrow x>9\)
\(c,\sqrt{P}_{min}=0\Rightarrow\frac{\sqrt{x}-3}{4}=0\)
\(\Leftrightarrow\sqrt{x}-3=0\Rightarrow\sqrt{x}=3\Rightarrow x=9\)
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
