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bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
học lớp 9 chưa mà đòi đăng ? :))
a) Ta có : \(A=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)
Để A nhận giá trị = 0 thì \(\sqrt{x}=0\)<=> x = 0 ( tmđk )
Vậy với x = 0 thì A = 0
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) P = B : A = \(\frac{\frac{\sqrt{x}}{\sqrt{x}+3}}{\frac{\sqrt{x}}{\sqrt{x}-5}}=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}}{\sqrt{x}+3}\times\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
Xét hiệu P - 1 ta có :
\(\frac{\sqrt{x}-5}{\sqrt{x}+3}-1=\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{-8}{\sqrt{x}+3}\)
Vì \(\hept{\begin{cases}-8< 0\\\sqrt{x}+3>0\end{cases}}\Rightarrow\frac{-8}{\sqrt{x}+3}< 0\)hay P - 1 < 0
=> P < 1
a) \(A=0\Rightarrow\frac{x+5\sqrt{x}}{x-25}=0\Rightarrow x+5\sqrt{x}=0\Leftrightarrow x=0\)(thỏa mãn).
b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)
\(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2\sqrt{x}\left(\sqrt{x}+3\right)-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{2x+6\sqrt{x}-x-9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(B=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+3}\)
c) \(P=B\div A=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}=1-\frac{8}{\sqrt{x}+3}< 1\)
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
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Mk làm từng câu nhé !
a)\(A=\frac{x-\sqrt{x}}{x-1}\left(đk:x\ge0,x\ne1\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)(vì \(x\ge0\))
\(=\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(B=\frac{x-4}{x+2\sqrt{x}}\left(đk:x>0,x\ne4\right)\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=1+\frac{2}{\sqrt{x}}\)
a.\(DK:x\ge0,x\ne1\)
\(A=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
\(DK:x\ge0\)
\(B=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}}\)
b.\(A-B=\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{x-x+\sqrt{x}+2}{x+\sqrt{x}}=\frac{\sqrt{x}+2}{x+\sqrt{x}}>0\)
\(\Rightarrow A-B>0\Rightarrow A>B\)
c.Ta co:\(A.B=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+1}=1-\frac{3}{\sqrt{x}+1}\)
De \(A.B\in Z\)
\(\Rightarrow1-\frac{3}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\frac{3}{\sqrt{x}+1}\in Z\)
\(\Rightarrow3⋮\sqrt{x}+1\)
\(\Rightarrow x=4\)
d.Ta co: \(A.B=\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{2}\)
\(\Leftrightarrow2\sqrt{x}-4< \sqrt{x}+1\)
\(\Leftrightarrow x< 25\)
Để làm 1 lượt luôn, hihi !
b)
Có \(A=\frac{\sqrt{x}}{\sqrt{x}+1}\)\(< 1\)(vì \(x\ge0\))
\(B=1+\frac{2}{\sqrt{x}}>1\)(Vì \(x\ge0\))
=>A<B
c)Ta có:
A.B=\(\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{2}{\sqrt{x}}=\frac{\sqrt{x}+2}{\sqrt{x}+1}=1+\frac{1}{\sqrt{x}+1}\)
Để:\(A.B\in\frac{Z\Rightarrow1}{\sqrt{x}+1}\in Z\)
Với:\(x\ge0\)có \(\orbr{\begin{cases}\sqrt{x}\in Z\\\sqrt{x}\notin Z\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x}+1\in Z\\\sqrt{x}+1\notin Z\end{cases}\Leftrightarrow}\orbr{\begin{cases}\frac{1}{\sqrt{x}+1}\in Z\left(tm\right)\\\frac{1}{\sqrt{x}+1}\notin Z\left(ktm\right)\end{cases}}}\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;-2\right\}\)
Mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=0\Leftrightarrow x=0\left(\frac{t}{m}\right)\)
Vậy: để A.B thuộc Z <=> x=0
d)Có A.B < 1/2
<=>\(1+\frac{1}{\sqrt{x}+1}< \frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{x}+1}+\frac{1}{2}< 0\)(vô lý)
Vậy không có x t/m A.B < 1/2