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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.
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 ..............................................................................................................................
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
1.
a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b,
\(M=(\dfrac{\sqrt{x}}{\sqrt{x}-2}\times\dfrac{\sqrt{x}}{\sqrt{x}+2})\times\dfrac{x-4}{\sqrt{4x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2+\sqrt{x}-2\right)}{x-4}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=(\sqrt{x}\times2\sqrt{x})\times\dfrac{1}{2\sqrt{x}}\)
\(=\sqrt{x}\)
c,
\(M>3\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Bài 2:
a: \(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}=\sqrt{1-a}\)
b: Để A=căn A thì A=1 hoặc A=0
=>A=1
=>1-a=1
=>a=0
c: Thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\) vào A, ta được:
\(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4x+2\sqrt{x}-4}{\sqrt{x}^2-2^2}\right):\left(\dfrac{2}{2-\sqrt{x}}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(\sqrt{x}-2\right)-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(P=\left(\dfrac{-4-4\sqrt{x}-x+x-2\sqrt{x}-4x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\right)\)
\(P=\dfrac{-4x\left(\sqrt{x}\left(2-\sqrt{x}\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4x\left(-\sqrt{x}\left(\sqrt{x}-2\right)\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{\sqrt{16x^3}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
Có j bạn xem lại coi có sai xót chỗ nào ko nhé, mk ko chắc là đúng 100% đâu.
Câu 3:
\(C=\dfrac{3\sqrt{x}-x+x+9}{9-x}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
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) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
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
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