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1: Sửa đề: \(B=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
2: Để B<=-1/2 thì B+1/2<=0
=>-3/căn x+3+1/2<=0
=>-6+căn x+3<=0
=>căn x<=3
=>0<x<9
3: Để B là số nguyên thì \(\sqrt{x}+3=3\)
=>x=0
ĐK: x>0,x\(\ne4\)
a) Ta thay x=\(\dfrac{1}{4}\) vào \(A=\dfrac{6}{x+2\sqrt{x}}=\dfrac{6}{\dfrac{1}{4}+2\sqrt{\dfrac{1}{4}}}=\dfrac{6}{\dfrac{1}{4}+2.\dfrac{1}{2}}=\dfrac{6}{\dfrac{1}{4}+1}=6:\left(\dfrac{1}{4}+1\right)=6:\dfrac{5}{4}=6.\dfrac{4}{5}=\dfrac{24}{5}=4,8\)B=\(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{x-4}-\dfrac{2}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2\sqrt{x}-4+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}=\dfrac{6}{4-x}\)
b) Ta có M=\(\dfrac{A}{B}=A\div B=\dfrac{6}{x+2\sqrt{x}}\div\dfrac{6}{4-x}=\dfrac{6}{x+2\sqrt{x}}.\dfrac{4-x}{6}=\dfrac{4-x}{x+2\sqrt{x}}=\dfrac{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{2-\sqrt{x}}{\sqrt{x}}\)
Ta lại có M>1\(\Leftrightarrow\dfrac{2-\sqrt{x}}{\sqrt{x}}>1\Leftrightarrow2-\sqrt{x}>\sqrt{x}\Leftrightarrow2>2\sqrt{x}\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Kết hợp với ĐK
Vậy 0<x<1 thì M>1
c) Ta có M\(=\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{2}{\sqrt{x}}-1\)
Vậy để \(M\in Z\) thì \(\sqrt{x}\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)
Vì \(\sqrt{x}>0\)
Nên \(\sqrt{x}\in\left\{1;2\right\}\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(ktm\right)\end{matrix}\right.\)
Vậy x=1 thì M\(\in Z\)
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a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
a: Để P là số nguyên thì \(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{9;1;16;0\right\}\)
b: Để P là só nguyên thì \(2\sqrt{x}+6-7⋮\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\{1;-1;7;-7\right\}\)
=>căn x+3=7
=>căn x=4
=>x=16
c: Để P là số nguyên thì \(3\sqrt{x}-1⋮2\sqrt{x}+1\)
\(\Leftrightarrow6\sqrt{x}-2⋮2\sqrt{x}+1\)
=>\(6\sqrt{x}+3-5⋮2\sqrt{x}+1\)
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
=>x=0 hoặc x=4
Áp dụng BĐT Cauchy , ta có :
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)
\(\dfrac{y^2}{\sqrt{1-y^2}}=\dfrac{y^3}{y\sqrt{1-y^2}}\ge\dfrac{y^3}{\dfrac{y^2+1-y^2}{2}}=2y^3\)
\(\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{z^3}{z\sqrt{1-z^2}}\ge\dfrac{z^3}{\dfrac{z^2+1-z^2}{2}}=2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\)
1) điều kiện \(x\ge0;x\ne\dfrac{1}{49}\)
\(Q=\dfrac{\sqrt{x}+4}{1-7\sqrt{x}}+\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{24\sqrt{x}}{7x+6\sqrt{x}-1}\)
\(Q=\dfrac{-\sqrt{x}-4}{7\sqrt{x}-1}+\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(Q=\dfrac{\left(-\sqrt{x}-4\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}-2\right)\left(7\sqrt{x}-1\right)+24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(Q=\dfrac{-x-\sqrt{x}-4\sqrt{x}-4+7x-\sqrt{x}-14\sqrt{x}+2+24\sqrt{x}}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(Q=\dfrac{6x+4\sqrt{x}-2}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(6\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(7\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}\)
\(1.\) 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....
\(\dfrac{1}{P}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{2.\left(\sqrt{x}+1\right)-2}{\sqrt{x}+1}=2-\dfrac{2}{\sqrt{x}+1}\)(\(x>0;\sqrt{x}+1>1\))
\(\dfrac{1}{P}\in Z\Leftrightarrow\dfrac{2}{\sqrt{x}+1}\in Z\Leftrightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\) (do mẫu thức lớn hơn 1 nên có thế làm theo cách này)theo điều kiện ta chỉ có 1 TH:
\(\sqrt{x}+1=2\Leftrightarrow x=1\left(TM\right)\)
vậy.............
\(P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
\(\Rightarrow\dfrac{1}{P}=\dfrac{2\sqrt{x}}{\sqrt{x}+1}=\dfrac{2\sqrt{x}+2-2}{\sqrt{x}+1}=2-\dfrac{2}{\sqrt{x}+1}\)
Để \(\dfrac{1}{P}\) nguyên
\(\Leftrightarrow\sqrt{x}+1\in\text{Ư}\left(2\right)\)
Ta có bảng :