Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Bài 2:
a: \(P=\dfrac{a-1}{2\sqrt{a}}\cdot\left(\dfrac{\sqrt{a}\left(a-2\sqrt{a}+1\right)-\sqrt{a}\left(a+2\sqrt{a}+1\right)}{a-1}\right)\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}=-2\sqrt{a}\)
b: Để P>=-2 thì P+2>=0
\(\Leftrightarrow-2\sqrt{a}+2>=0\)
=>0<=a<1
Bài 1:
\(a,E=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(b,E>0\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}}>0\)
Mà: \(\sqrt{x}>0\\ \Rightarrow\sqrt{x}-1>0\\ \Leftrightarrow\sqrt{x}>1\\ \Leftrightarrow x>1\)
Bài 2:
\(a,G=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\left(\sqrt{x}+1\right)\\ =\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)\\ =\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}\\ =\sqrt{x}-1\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
a)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ P=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b)
\(Q< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}< 0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>0\\x< 4\end{matrix}\right.\\ \Leftrightarrow0< x< 4\)
a) điều kiện để M có nghĩa là \(x\ge0;x\ne4\)
b) \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{x-4}\right):\dfrac{1}{x-4}\)
\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(M=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(M=\dfrac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{1}\)
\(M=\dfrac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{1}=\dfrac{-1}{1}=-1\)
\(a.A=\left(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\dfrac{1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{1+\sqrt{3}-1-\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}\right):\dfrac{1}{\sqrt{3}}\)
\(A=\left(\dfrac{0}{1-3}\right):\dfrac{1}{\sqrt{3}}\) \(=0:\dfrac{1}{\sqrt{3}}=0\)
b. B được xác định ⇔ x > 0 ; \(x\ne1\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}-1}{x-\sqrt{x}}\)
\(B=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\).
\(B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c. Giả Sử A = \(\dfrac{1}{6}B\)
⇔ 0 = \(\dfrac{1}{6}\times\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
⇔ 0=\(\dfrac{\sqrt{x}-1}{6\sqrt{x}}\)
⇔0 = \(\sqrt{x}-1\)
⇔x = 1(không thỏa mãn)
⇒ A ≠ \(\dfrac{1}{6}B\)
Vậy A ≠ \(\dfrac{1}{6}B\) (Do x không có giá trị nào thỏa mãn)
a) A=\(\dfrac{\sqrt{x}[\left(\sqrt{x}\right)^3-1]}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\) A=\(\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2\)
A=\(x-\sqrt{x}+1\)
b) A=\(\dfrac{3}{4}\)
=> \(x-\sqrt{x}+1=\dfrac{3}{4}\)
\(x-\sqrt{x}+\dfrac{1}{4}=0\)
\(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\)
=> \(\sqrt{x}=\dfrac{1}{2}\)
=> \(x=\dfrac{1}{4}\)
a/ ĐKXĐ: \(x\ge0;x\ne1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{x+1-2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
= \(\dfrac{x+1+\sqrt{x}}{x+1}:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(x+1\right)\left(\sqrt{x}-1\right)}-1\)
\(=\dfrac{\left(x+1+\sqrt{x}\right)\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(x+1\right)\left(\sqrt{x}-1\right)^2}-1\)
= \(\dfrac{x+1+\sqrt{x}}{\sqrt{x}-1}-1=\dfrac{x+2}{\sqrt{x}-1}\)
b/ Ta có:
\(Q=P-\sqrt{x}\)
= \(\dfrac{x+2}{\sqrt{x}-1}-\sqrt{x}\)
= \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để Q nhận giá trị nguyên thì \(1+\dfrac{3}{\sqrt{x}-1}\in Z\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}-1}\in Z\) ( vì 1\(\in Z\) )
\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\\\sqrt{x}-1=-3\\\sqrt{x}-1=1\\\sqrt{x}-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=-2\\\sqrt{x}=2\\\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\left(tm\right)\\\\x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy để biểu thức \(Q=P-\sqrt{x}\) nhận giá trị nguyên thì x=\(\left\{16;4;0\right\}\)
\(a,ĐK:x>0;x\ne1\\ b,M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}-1\\ c,M< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0< x< 1\)