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Bài 1:
a)Với x > 0;x ≠ 4 ta có:
\(\left(\dfrac{1}{x-4}-\dfrac{1}{x+4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}\)
\(=\dfrac{1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}+2\right)-\dfrac{1}{\left(\sqrt{x}+2\right)^2}\cdot\left(\sqrt{x}+2\right)\)
\(=\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}+2\right)-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4}{x-4}\)
c)\(\left(\dfrac{\sqrt{b}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\dfrac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\cdot\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)=b-a\)
Bài 2:
a)Với a > 0;a ≠ 1;a ≠ 2 ta có
\(P=\left(\dfrac{\sqrt{a}^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a}^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}=\dfrac{2\left(a-2\right)}{a+2}\)
b)Ta có:
\(P=\dfrac{2\left(a-2\right)}{a+2}=\dfrac{2a-4}{a+2}=\dfrac{2\left(a+2\right)-8}{a+2}=2-\dfrac{8}{a+2}\)
P nguyên khi \(2-\dfrac{8}{a+2}\) nguyên⇒\(\dfrac{8}{a+2}\) nguyên⇒\(a+2\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(TH1:a+2=1\Rightarrow a=-1\left(loai\right)\)
\(TH2:a+2=-1\Rightarrow a=-3\left(loai\right)\)
\(TH3:a+2=2\Rightarrow a=0\left(loai\right)\)
\(TH4:a+2=-2\Rightarrow a=-4\left(loai\right)\)
\(TH5:a+2=4\Rightarrow a=2\left(loai\right)\)
\(TH6:a+2=-4\Rightarrow a=-6\left(loai\right)\)
\(TH7:a+2=8\Rightarrow a=6\left(tm\right)\)
\(TH8:a+2=-8\Rightarrow a=-10\left(loai\right)\)
Vậy a = 6
1) +) ta có : \(C-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=\dfrac{3\sqrt{x}-x+\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{-\left(x-4\sqrt{x}+4\right)+3}{3\left(x+\sqrt{x}+1\right)}=\dfrac{-\left(\sqrt{x}-2\right)^2+3}{3\left(x+\sqrt{x}+1\right)}\)
không thể cm được đâu bn --> xem lại đề
2) +) ta có : \(D=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)
--> để \(D\in Z\Leftrightarrow\sqrt{x}+2\) là ước của 3 \(\Leftrightarrow\sqrt{x}+2\in\left\{\pm1;\pm3\right\}\)
\(\Leftrightarrow x=1\) vậy \(x=1\)
3) +) tương tự 2)
4) a) +) điều kiện xác định : \(x>0;x\ne4\)
ta có : \(A=\left(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}}\right):\dfrac{\sqrt{x}-2}{x+3\sqrt{x}}\)
\(\Leftrightarrow A=\left(\dfrac{2\sqrt{x}-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}}{\sqrt{x}-2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\)
b) ta có : \(A=3\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}=3\Leftrightarrow\sqrt{x}-3=3\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}=3\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\) vậy \(x=\dfrac{9}{4}\)
c) ta có : \(B=A.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}.\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{x-9}{x-4}=1-\dfrac{5}{x-4}\)
tương tự 2 )
\(\)
1. a) \(A=\left(\dfrac{\sqrt{x}-1+x-\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}\right).\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)ĐK x\(\ne\)0,1
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(x-1\right)2\sqrt{x}}{\left(x-\sqrt{x}\right)\left(x-1\right)}=\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
b) A<-1 <=> \(\dfrac{2\sqrt{x}}{x-\sqrt{x}}< -1\)\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}+1< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+x-\sqrt{x}}{x-\sqrt{x}}< 0\)\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x-\sqrt{x}}< 0\)
\(\Leftrightarrow x-\sqrt{x}< 0\) (vì \(x+\sqrt{x}>0\left(\forall x>0\right)\))
\(\Leftrightarrow x< \sqrt{x}\Leftrightarrow x^2< x\Leftrightarrow x^2-x< 0\)
\(\Leftrightarrow x\in\left(0;1\right)\Leftrightarrow0< x< 1\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Thay \(x=5-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{-5\left(\sqrt{3}-\sqrt{2}\right)+2}{\sqrt{3}-\sqrt{2}+3}=\dfrac{-5\sqrt{3}+5\sqrt{2}+2}{\sqrt{3}-\sqrt{2}+3}\simeq0,124\)
d: Để A=1/2 thì \(\sqrt{x}+3=-10\sqrt{x}+4\)
\(\Leftrightarrow11\sqrt{x}=1\)
hay x=1/121
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a) ĐKXĐ: \(x\ge0;x\ne9\) . Rút gọn: \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x-2\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{x+\sqrt{x}-3\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{\sqrt{x}-3}-\dfrac{x-4\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-3\sqrt{x}-2\sqrt{x}+6+x+\sqrt{x}+3\sqrt{x}+3-x+4\sqrt{x}-7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
A>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)>-1\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+1>0\Leftrightarrow\dfrac{\sqrt{x}+2+\sqrt{x}-3}{\sqrt{x}-3}>0\Leftrightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{x}-1>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{x}-1< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>0,5\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 0,5\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0,25\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0,25\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow}}\left[{}\begin{matrix}x>9\\0\le x< 0,25\end{matrix}\right.\)
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 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
2. \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\)
\(\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)
3. Ta có: VT=\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}:\sqrt{a}\right).\left(\dfrac{1-\sqrt{a}}{1-a}\right)=\left[\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}.\dfrac{1}{\sqrt{a}}\right].\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]=\dfrac{1+\sqrt{a}+a}{\sqrt{a}}.\dfrac{1}{1+\sqrt{a}}=\dfrac{1+\sqrt{a}+a}{\sqrt{a}+a}=\dfrac{1}{\sqrt{a}+a}+1\)
??? Sao rút gọn rồi ra kì vậy nhờ =="
1,
a.
\(\left[{}\begin{matrix}x-5\sqrt{x}+6\ne0\\\sqrt{x}-2\ne0\\3-\sqrt{x}\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\ne0\\\sqrt{x}\ne2\\\sqrt{x}\ne3\end{matrix}\right.\)
\(\left[{}\begin{matrix}\sqrt{x}\ne3\\\sqrt{x}\ne2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
Vậy ĐKXĐ : \(\left[{}\begin{matrix}x\ne9\\x\ne4\end{matrix}\right.\)
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)