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Lời giải:
ĐKXĐ:......
a) Ta có:
\(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}-\frac{4x}{x-9}=\frac{(3+\sqrt{x})^2-(3-\sqrt{x})^2}{(3-\sqrt{x})(3+\sqrt{x})}-\frac{4x}{x-9}\)
\(=\frac{9+x+6\sqrt{x}-(9+x-6\sqrt{x})}{9-x}-\frac{4x}{x-9}=\frac{-12\sqrt{x}}{x-9}-\frac{4x}{x-9}=\frac{-4\sqrt{x}(3+\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{4\sqrt{x}}{3-\sqrt{x}}\)
Và:
\(\frac{5}{3-\sqrt{x}}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{5\sqrt{x}}{3\sqrt{x}-x}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}\)
Do đó:
\(C=\frac{4\sqrt{x}}{3-\sqrt{x}}: \frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}=\frac{4\sqrt{x}}{3-\sqrt{x}}.\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}=\frac{4x}{\sqrt{x}-2}\)
b)
Nếu $C\leq 0$ thì \(|C|=-C\) (không thỏa mãn)
Nếu $C>0$ thì \(|C|=C>0>-C\) (thỏa mãn)
Vậy để \(|C|> -C\) thì \(C>0\Leftrightarrow \frac{4x}{\sqrt{x}-2}>0\Leftrightarrow \sqrt{x}-2>0\) (do \(x>0)\)
\(\Leftrightarrow x> 4\)
Kết hợp đkxđ suy ra điều kiện của $x$ là \(x>4; x\neq 9\)
c)
\(C^2=40C\Leftrightarrow C(C-40)=0\Leftrightarrow \left[\begin{matrix} C=0\\ C=40\end{matrix}\right.\)
Nếu $C=0$ thì \(\frac{4x}{\sqrt{x}-2}=0\Rightarrow x=0\) (không t/m ĐKXĐ)
Nếu \(C=40\Leftrightarrow \frac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10(\sqrt{x}-2)\)
\(\Rightarrow \sqrt{x}=5\pm \sqrt{5}\Rightarrow x=(5\pm \sqrt{5})^2\)
Na: cái này là giải pt bậc 2 đơn giản thôi bạn:
\(x=10(\sqrt{x}-2)\)
\(\Rightarrow x-10\sqrt{x}+20=0\)
\(\Rightarrow (\sqrt{x}-5)^2-5=0\Rightarrow (\sqrt{x}-5)^2=5\)
\(\Rightarrow \sqrt{x}-5=\pm \sqrt{5}\Rightarrow \sqrt{x}=5\pm \sqrt{5}\) đó bạn.
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 \(x=3\)
a: \(P=\dfrac{9x+6\sqrt{x}+1-9x+6\sqrt{x}-1+4x}{9-x}:\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{4x+12\sqrt{x}}{9-x}\cdot\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\)
\(=\dfrac{4x}{\sqrt{x}-2}\)
b: Để P^2=40P thì P(P-40)=0
=>P=0(loại) hoặc P=40
=>4x=40 căn x-80
=>4x-40 căn x+80=0
=>x-10 căn x+20=0
=>căn x=5+căn 5 hoặc căn x=5-căn 5
=>x=30+10 căn 5 hoặc x=30-10 căn 5
a: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\frac{1}{x+1}\right).\frac{x+1}{\sqrt{x}-1}\)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 ....
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 )
a) Ta có:
\(P=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{2\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{x-9}\right):\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\left(\frac{2x-6}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3x+3}{x-9}\right):\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2x-6+x+3\sqrt{x}-3x-3}{x-9}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+3}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\)
b) \(P< \frac{-1}{2}\Rightarrow\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}< \frac{-1}{2}\)
.....Chưa nghĩ ra....
c) Ta có: \(\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)^2}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}-3=0\Rightarrow x=9\)
Vậy Min P = 0 khi x =9.
k - kb với tớ nhia mn!
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
điều kiện xác định : \(x>0;x\ne9\)
a) ta có : \(C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}-\dfrac{4x}{x-9}\right):\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{3\sqrt{x}-x}\right)\)
\(\Leftrightarrow C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}+\dfrac{4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}+\dfrac{4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{\left(3+\sqrt{x}\right)^2-\left(3-\sqrt{x}\right)^2+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{12\sqrt{x}+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right)\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\right)\) \(\Leftrightarrow C=\left(\dfrac{4\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right)\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\right)=\dfrac{4x}{\sqrt{x}-2}\)b) để \(\left|C\right|>-C\) \(\Leftrightarrow C< 0\) \(\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}< 0\) \(\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
c) để \(C^2=40C\Leftrightarrow C^2-40C=0\Leftrightarrow C\left(C-40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}C=0\\C=40\end{matrix}\right.\)
+) \(C=0\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=0\) \(\Leftrightarrow x=0\left(loại\right)\)
+) \(C=40\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10\sqrt{x}-20\)
\(\Leftrightarrow x-10\sqrt{x}+20=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5+3\sqrt{5}\left(N\right)\\\sqrt{x}=5-3\sqrt{5}\left(L\right)\end{matrix}\right.\)
ta có : \(\sqrt{x}=5+3\sqrt{5}\Leftrightarrow x=70+30\sqrt{5}\)
vậy ..............................................................................................................................
Mysterious Person giúp e