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a) ĐKXĐ: \(x\ge0,x\ne1\)
b) \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
c) \(A=2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\)(vô lý do \(2\sqrt{x}\ge0\forall x\))
Vậy \(S=\varnothing\)
Bài 3:
\(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt[]{x}+1}\\ DKXD:x\ne1;x\ge0\\ A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\\ A=\sqrt{x}-1+\sqrt{x}\\ A=2\sqrt{x}+1\)
\(C.A< -1\Leftrightarrow2\sqrt{x}-1< -1\\ \Leftrightarrow2\sqrt{x}< 0\\ \Leftrightarrow x< 0\left(ktmdk\right)\\ =>BPTVN:S=\varnothing\)
a) ĐK: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) \(A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)c) Ta có A<-1\(\Leftrightarrow2\sqrt{x}-1< -1\Leftrightarrow2\sqrt{x}< 0\Leftrightarrow\sqrt{x}< 0\left(ktm\right)\)
Vậy không có giá trị của x để A<-1
a.ĐK:
\(\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\\\sqrt{x}xđ\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b.\(A=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}.\)
c.Có :\(2\sqrt{x}\ge0\) nên không có giá trị nào của x để A<-1.
\(1;2.A=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\left(x\ge0;x\ne1\right)\)
\(3.A< -1\Leftrightarrow2\sqrt{x}-1< -1\)
\(\Leftrightarrow2\sqrt{x}< 0\) ( Vô lý )
KL : Vậy , không có giá trị nào cua x để \(A< -1\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)
Bài 1:
a: \(B=\dfrac{\sqrt{x}+x+\sqrt{x}-x}{1-x}\cdot\dfrac{x-1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-3}\)
b: Để B=-1 thì \(2\sqrt{x}=-\sqrt{x}+3\)
=>3 căn x=3
=>căn x=1
hay x=1(loại)
Câu a : ĐKXĐ : \(x\ge0\) và \(x\ne4\)
Câu b : \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}+\dfrac{1}{2-\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-4-5-\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
Câu c :
\(A< 1\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}< 1\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)< \left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)\)
\(\Leftrightarrow x-\sqrt{x}-6< x+\sqrt{x}-6\)
\(\Leftrightarrow-2\sqrt{x}< 0\) ( Luôn đúng với mọi x khi \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\))
Vậy các giá của x để A < 1 là \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Thay x=9 vào A, ta được:
\(A=\dfrac{3-1}{3+1}=\dfrac{1}{2}\)
c: Ta có: P=AB
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}-1}+\dfrac{5-x}{x-1}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\left(\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{6}{\sqrt{x}+1}\)
a: ĐKXĐ: x>=0; x<>1
b: Khi x=9 thì \(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{x+2\sqrt{x}-3+4\sqrt{x}+4+5-x}{x-1}\)
\(=\dfrac{6\sqrt{x}+6}{\left(\sqrt{x}+1\right)^2}=\dfrac{6}{\sqrt{x}+1}\)
a: ĐKXĐ: x>=0; x<>1
b: \(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
c: Để A<-1 thì A+1<0
\(\Leftrightarrow2\sqrt{x}< 0\)
hay \(\sqrt{x}< 0\)(vô lý)