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\(a.A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}=\dfrac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2-5\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
\(\left(x\ge0;x\ne1\right)\)
\(b.A=\dfrac{1}{2}\Leftrightarrow\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{4-10\sqrt{x}-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}=0\)
\(\Leftrightarrow-11\sqrt{x}+1=0\)
\(\Leftrightarrow x=\dfrac{1}{121}\left(TM\right)\)
KL...........
a: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\dfrac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
c: Sửa đề:\(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y}-1}{\left(x-1\right)^2}=\dfrac{1}{\left(x-1\right)}\)
\(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\)
\(\Leftrightarrow3< 1\) ( Vô lý )
\(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\)
\(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\)
\(\Leftrightarrow a-2\sqrt{ab}+b< a-b\)
\(\Leftrightarrow2b-2\sqrt{ab}< 0\)
\(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\)
Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\)
\(\RightarrowĐpcm.\)
\(2a.\) Áp dụng BĐT Cauchy , ta có :
\(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\)
\(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\)
\(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\)
\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\)
\(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\)
Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được :
\(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\)
\(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\)
\(\Leftrightarrow x-4=a^2\)
\(\Leftrightarrow x=a^2+4\left(TM\right)\)
\(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\)
\(\Leftrightarrow x+4=x^2+4x+4\)
\(\Leftrightarrow x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
KL....
\(A=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}.\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
ĐKXĐ: \(x\ge0,x\ne1\)
\(A=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(A=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(A=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}=\dfrac{x+2}{\sqrt{x}^3-1^3}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{x+2+x-1-1}{\sqrt{x}^3-1}=\dfrac{2x}{\sqrt{x}^3-1}\)
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2+2+1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)+3}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}\)
\(=1+\dfrac{3}{\sqrt{x}-2}\)
Để A nguyên thì \(\dfrac{3}{\sqrt{x}-2}\) phải nguyên
Do đó 3⋮( \(\sqrt{x}\) -2)
⇒\(\sqrt{x}-2\) ∈ Ư(3)
Mà Ư(-1;1;-3;3)
Nên \(\left[{}\begin{matrix}\sqrt{x}-2=-1\\\sqrt{x}-2=1\\\sqrt{x}-2=-3\\\sqrt{x}-2=3\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=3\\\sqrt{x}=-1\left(v\text{ô}l\text{í}\right)\\\sqrt{x}=5\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=1\left(nh\text{ận}\right)\\x=9\left(nh\text{ận}\right)\\x=25\left(nh\text{ận}\right)\end{matrix}\right.\)
Vậy x=1 hay x=9 hay x=25 thì A nguyên