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\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
b) Xét hiệu:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)
\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)
\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)
Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\)
\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)
Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)
(giúp cậu nó nha)
\(a,B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\left(x\ge0;x\ne1\right)\\ B=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ B=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
b: Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)
\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)
ĐKXĐ: x>=0; x<>1
a: \(B=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left(\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\right)\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\sqrt{x}+1}:\left[\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2\right]\)
\(=\dfrac{\sqrt{x}\left(x-1\right)^2}{\left(x-1\right)^2\cdot\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
b: Khi x=4-2căn 3=(căn 3-1)^2 thì \(B=\dfrac{\sqrt{3}-1}{\sqrt{3}-1+1}=\dfrac{\sqrt{3}-1}{\sqrt{3}}=\dfrac{3-\sqrt{3}}{3}\)
c: B=2/3
=>căn x/căn x+1=2/3
=>căn x=2
=>x=4
d: \(B-1=\dfrac{\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{1}{\sqrt{x}+1}< 0\)
=>B<1
e: B>1
=>-1/căn x+1>0
=>căn x+1<0(vô lý)
=>KO có x thỏa mãn
f: B nguyên khi căn x chia hết cho căn x+1
=>căn x+1-1 chia hết cho căn x+1
=>căn x+1=1 hoặc căn x+1=-1(loại)
=>căn x=0
=>x=0
Đk:\(x>0;x\ne1\)
\(B=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
\(B=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)\(\Leftrightarrow x=9\) (tm)
Vậy..
a) \(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)
\(B=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(B=\dfrac{1}{\sqrt{x}-1}\)
b) Với \(B=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\)
Vậy...
Chúc bạn học tốt
1: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)
2: \(P=A\cdot B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(\Leftrightarrow P-2=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}>0\)
=>P>2
a) Ta có: \(A=\dfrac{3+2\sqrt{3}}{\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=2+\sqrt{3}-\sqrt{3}-\sqrt{2}+\sqrt{2}\)
=2
Ta có: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
a) Ta có: \(B=\left(\dfrac{3}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3\sqrt{x}-6-\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}-8}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để \(B=\dfrac{1}{3}\) thì \(\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+2\)
\(\Leftrightarrow2\sqrt{x}=2\)
\(\Leftrightarrow x=1\)(thỏa ĐK)
a) B= \(\left(\dfrac{3\left(\sqrt{x}-2\right)-1\left(\sqrt{x}+2\right)}{x-4}\right):\left(\dfrac{\sqrt{x}-6+1\left(\sqrt{x}-2\right)}{x-2\sqrt{x}}\right)\)
\(=\dfrac{2\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\sqrt{x}}{2\sqrt{x}-8}\)=\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
b) Để B=\(\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}=\dfrac{1}{3}\Leftrightarrow\sqrt{x}+2=3\sqrt{x}\Rightarrow x=1\)
a) ĐKXĐ: \(x>0,x\ne1\)
\(B=1:\dfrac{\left(x+2\right)\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\left(x-1\right)-\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)
b) \(B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)
Áp dụng BĐT Cauchy cho 2 só dương:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\dfrac{\sqrt{x}.1}{\sqrt{x}}}=2\)
\(\Rightarrow B=1+\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge1+2=3\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)