Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\\ b,x=36\Leftrightarrow A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Leftrightarrow3\sqrt{x}=2-\sqrt{x}\\ \Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\\ d,A\in Z\Leftrightarrow1+\dfrac{2}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\\ \Leftrightarrow x\in\left\{0;1;9;16\right\}\)
\(e,A:B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}=-2\\ \Leftrightarrow\sqrt{x}=-2\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{3}\left(ktm\right)\\ \Leftrightarrow x\in\varnothing\)
a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)
Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)
b, Với \(x\ge0;x\ne4\)
\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)
\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )
c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay
\(\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{2+\sqrt{x}}{\sqrt{x}-2}=\frac{3x}{\sqrt{x}-2}\)
\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)
\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))
a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)
Thay vào biểu thức A ta được : \(\frac{1}{2-1}=1\)
b, Với \(x\ge0;x\ne1\)
\(Q=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{x-1}-1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2-x+1}{x-1}\)
\(=\frac{x+\sqrt{x}-2-x+1}{x-1}=\frac{\sqrt{x}-1}{x-1}=\frac{1}{\sqrt{x}+1}\)
c, Ta có : \(\frac{1}{Q}+P\le4\)hay\(1:\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\le4\)ĐK : \(x\ne1\)
\(\Leftrightarrow\frac{x-1+1}{\sqrt{x}-1}-4\le0\Leftrightarrow\frac{x-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)
\(\Leftrightarrow\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\le0\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)do \(\left(\sqrt{x}-2\right)^2\ge0\)
Kết hợp với đk, vậy \(x< 1\)
1, thay x=4 (TMĐKXĐ) vào P ta được:
P=\(\dfrac{1}{\sqrt{4}-1}\)=1
vậy khi x=4 thì P =1
2,với x≥0,x≠1:
Q=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}\)-\(\dfrac{2}{\sqrt{x}-1}-1\)=\(\dfrac{\sqrt{x}-2-\sqrt{x}+1}{\sqrt{x}-1}\)=\(\dfrac{-1}{\sqrt{x}-1}\)
vậy Q=\(\dfrac{-1}{\sqrt{x}-1}\)
3,\(\dfrac{1}{Q}+P\le4\)
⇒1/\(\dfrac{-1}{\sqrt{x}-1}\)+\(\dfrac{1}{\sqrt{x}-1}\)≤4⇔\(\dfrac{-\sqrt{x}-1}{1}+\dfrac{1}{\sqrt{x}-1}\le4\)⇔\(\dfrac{-x+1+1}{\sqrt{x}-1}-4\le0\)⇔\(\dfrac{-x+2-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)⇔\(\dfrac{-x-4\sqrt{x}+6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}-6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}+4-10}{\sqrt{x}-1}\le0\)
\(\dfrac{ \left(\sqrt{x}+2\right)^2-10}{\sqrt{x}-1}\le0\)⇒\(\sqrt{x}-1\le0\) (vì (\(\sqrt{x}+2\))\(^2\)≥0 ∀ x hay (\(\sqrt{x}+2\))\(^2\)-10>0 ∀ x)
⇔x≤1 (KTM)
vậy không có giá trị nào của x TM để \(\dfrac{1}{Q}+P\le4\)
a: \(A=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)