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a: \(P=\dfrac{a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}{a-4}\)
\(=\dfrac{4\sqrt{a}+8}{a-4}=\dfrac{4}{\sqrt{a}-2}\)
b: Khi a=1/9 thì \(P=\dfrac{4}{\dfrac{1}{3}-2}=4:\dfrac{-5}{3}=-\dfrac{12}{5}\)
c: Để P=2 thì \(2\sqrt{a}-4=4\)
=>2căn a=8
=>căn a=4
hay a=16
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
a) \(A=\left(\dfrac{1}{3}+\dfrac{3}{x^2-3x}\right):\left(\dfrac{x^2}{27-3x^2}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3\left(x^2-3x\right)}:\left(\dfrac{x^2}{3\left(9-x^2\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\left(\dfrac{x^2}{3.\left(3-x\right).\left(3+x\right)}+\dfrac{1}{x+3}\right)\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+3.\left(3-x\right)}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}:\dfrac{x^2+9-3x}{3.\left(3-x\right).\left(3+x\right)}\)
\(\Rightarrow A=\dfrac{x^2-3x+9}{3x.\left(x-3\right)}.\dfrac{3.\left(3x-x\right).\left(3+x\right)}{x^2+9-3x}\)
\(\Rightarrow A=\dfrac{1}{x.\left(x-3\right)}.\left(-\left(x-3\right)\right).\left(3+x\right)\)
\(\Rightarrow A=\dfrac{1}{x}.\left(-1\right).\left(3+x\right)\)
\(\Rightarrow A=-\dfrac{1}{x}.\left(3+x\right)\)
dù ko đáp ứng điều kiện nhưng có ai cấm làm đâu 
a) \(A=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-x}{\sqrt{x}-1}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}-\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\sqrt{x}+\sqrt{x}\right)\left(1+\dfrac{1}{\sqrt{x}}\right)\)
\(=2\sqrt{x}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)
b) A = 4 \(\Leftrightarrow\) \(2\sqrt{x}+2=4\)
\(\Leftrightarrow2\sqrt{x}=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
c) Để \(\dfrac{3}{A}\) là số nguyên thì \(3⋮A\)
hay A \(\in\) Ư(3) = {1;-1;3;-3}
*\(2\sqrt{x}+2=1\Leftrightarrow2\sqrt{x}=-1\)(loại)
*\(2\sqrt{x}+2=-1\Leftrightarrow2\sqrt{x}=-3\)(loại)
*\(2\sqrt{x}+2=3\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{\sqrt{2}}\)(nhận)
*\(2\sqrt{x}+2=-3\Leftrightarrow2\sqrt{x}=-5\) (loại)
Vậy x = \(\dfrac{1}{\sqrt{2}}\).
a: \(A=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right):\dfrac{x^2-4+10-x^2}{x+2}\)
\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}=\dfrac{-1}{x-2}\)
b: |x|=1/2 nên x=1/2 hoặc x=-1/2
Khi x=1/2 thì \(A=\dfrac{-1}{\dfrac{1}{2}-2}=\dfrac{-1}{-\dfrac{3}{2}}=1\cdot\dfrac{2}{3}=\dfrac{2}{3}\)
Khi x=-1/2 thì \(A=\dfrac{-1}{-\dfrac{1}{2}-2}=1:\dfrac{5}{2}=\dfrac{2}{5}\)
c: Để A=2 thì x-2=-1/2
hay x=3/2
a, \(\sqrt{b}\) tồn tại \(\Leftrightarrow b>0\)
\(\left\{{}\begin{matrix}\sqrt{b}+1\ne0\\\sqrt{b}-1\ne0\\b-1\ne0\end{matrix}\right.\Leftrightarrow b\ne1\)
Vậy B có nghĩa khi \(\left\{{}\begin{matrix}b>0\\b\ne1\end{matrix}\right.\)
b,
\(B=\dfrac{\sqrt{b}}{\sqrt{b}+1}-\dfrac{\sqrt{b}}{\sqrt{b}-1}-\dfrac{2}{b-1}\)
\(=\dfrac{\sqrt{b}}{\sqrt{b}+1}-\dfrac{\sqrt{b}}{\sqrt{b}-1}-\dfrac{2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}\)
\(=\dfrac{\sqrt{b}\left(\sqrt{b}-1\right)-\sqrt{b}\left(\sqrt{b}+1\right)-2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{b-\sqrt{b}-b-\sqrt{b}-2}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{-2\left(\sqrt{b}+1\right)}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}=\dfrac{-2}{\sqrt{b}-1}=\dfrac{2}{1-\sqrt{b}}\)
c,
\(B>1\Leftrightarrow2>1-\sqrt{b}\)
\(\Leftrightarrow2-\left(1-\sqrt{b}\right)=1+\sqrt{b}>0\) (luôn đúng với mọi b)
=> Với mọi b có ĐKXĐ là b khác 0 và b > 1 thì B > 1
a: \(M=\dfrac{a^2+a+1}{a^2+1}:\left(\dfrac{a}{a-1}-\dfrac{2a}{\left(a-1\right)\left(a^2+1\right)}\right)\)
\(=\dfrac{a^2+a+1}{a^2+1}:\dfrac{a^3+a^2-2a}{\left(a-1\right)\left(a^2+1\right)}\)
\(=\dfrac{a^2+a+1}{a^2+1}\cdot\dfrac{\left(a-1\right)\left(a^2+1\right)}{a\left(a+2\right)\left(a-1\right)}\)
\(=\dfrac{a^2+a+1}{a^2+2a}\)
Để M là số nguyên thì \(a^2+a+1⋮a^2+2a\)
\(\Leftrightarrow a^2+2a-a+1⋮a^2+2a\)
=>-a^2+a chia hết cho a^2+2a
=>-a^2-2a+3a chia hết cho a^2+2a
=>3a chia hết cho a^2+2a
=>3 chia hết cho a+2
=>\(a+2\in\left\{1;-1;3;-3\right\}\)
hay \(a\in\left\{-1;-3;-5\right\}\)
b: Để M=7 thì \(a^2+a+1=7a^2+14a\)
=>7a^2+14a-a^2-a-1=0
=>6a^2+13a-1=0
hay \(a=\dfrac{-13\pm\sqrt{193}}{12}\)
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
a: \(A=\dfrac{a\left(\sqrt{a}+1\right)}{a-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-1\right)}-\dfrac{a+1}{\sqrt{a}}\)
\(=\dfrac{a^2+a\sqrt{a}+\sqrt{a}-1-a^2+1}{\sqrt{a}\left(a-1\right)}\)
\(=\dfrac{a\left(\sqrt{a}+1\right)}{\sqrt{a}\left(a-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
b: Để M>2 thì M-2>0
\(\Leftrightarrow\dfrac{\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}-1}>0\)
\(\Leftrightarrow\dfrac{\sqrt{a}-2}{\sqrt{a}-1}< 0\)
=>1<a<4
c: Để M=-1 thì \(\sqrt{a}=-\sqrt{a}+1\)
=>a=1/4