Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
A)
Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )
\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)
\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)
Có:
\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)
\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)
B)
\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)
\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)
\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)
\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$
T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)
\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)
Ta thấy:
\(\sqrt{\dfrac{1-y}{y}}\times\sqrt{\dfrac{y}{1-y}}=1\left(const\right)\)
=> Ta có thể đặt \(\sqrt{\dfrac{1-y}{y}}=t\left(t\ge0\right)\)
\(\Rightarrow\sqrt{\dfrac{y}{1-y}}=\dfrac{1}{t}\)
~ ~ ~
\(\sqrt{\dfrac{1-y}{y}}=t\)
\(\Rightarrow\dfrac{1-y}{y}=t^2\)
\(\Leftrightarrow1-y=yt^2\)
\(\Leftrightarrow yt^2+y=1\)
\(\Leftrightarrow y\left(t^2+1\right)=1\)
\(\Leftrightarrow y=\dfrac{1}{t^2+1}\)
~ ~ ~
\(x=\dfrac{1}{2}\left(t-\dfrac{1}{t}\right)=\dfrac{t^2-1}{2t}\)
\(\Rightarrow x^2+1=\dfrac{\left(t^2-1\right)^2}{4t^2}+1=\dfrac{\left(t^2-1\right)^2+4t^2}{4t^2}=\dfrac{\left(t^2+1\right)^2}{4t^2}\)
\(\Rightarrow\sqrt{x^2+1}=\left|\dfrac{t^2+1}{2t}\right|=\dfrac{t^2+1}{2t}\left(t\ge0\right)\)
~ ~ ~
\(B=\dfrac{2y\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\)
\(=\dfrac{2\times\dfrac{1}{t^2+1}\times\dfrac{t^2+1}{2t}}{\dfrac{t^2+1}{2t}-\dfrac{t^2-1}{2t}}\)
\(=\dfrac{\dfrac{1}{t}}{\dfrac{2}{2t}}=1\)
a) \(\sqrt{\dfrac{9x^2}{25}}+\dfrac{1}{5}x\) (x<0)
=\(\dfrac{-3x}{5}+\dfrac{x}{5}\) (vì x<0)
=\(\dfrac{-2x}{5}\)
b)2xy\(\sqrt{\dfrac{9x^2}{y^6}}-\sqrt{\dfrac{49x^2}{y^2}}\) (x<0 , y>0)
=2xy\(\dfrac{-3x}{y^3}+\dfrac{7x}{y}\)(vì x<y<0)
=\(\dfrac{-6x}{y^2}+\dfrac{7xy}{y^2}\)
=\(\dfrac{7xy-6x}{y^2}\)
c) \(\dfrac{1}{ab}\sqrt{a^6\left(a-b\right)^2}\) (a<b<0)
=\(\dfrac{1}{ab}\sqrt{a^6}\sqrt{\left(a-b\right)^2}\)
=\(\dfrac{1}{ab}\left(-a^3\right)\left(b-a\right)\) (vì a<b<0)
=\(\dfrac{\left(a-b\right)a^3}{a-b}\)
=a3
Cảm ơn bạn Thu Trang nhiều nhé, sau này có gì giúp đỡ nhau nha. 
Lời giải:
\(P=\frac{x+2}{(\sqrt{x})^3-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}\)
\(\frac{x+2}{\sqrt{x^3}-1}+\frac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\sqrt{x^3}-1}+\frac{x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{2x+1}{\sqrt{x^3}-1}-\frac{1}{\sqrt{x}-1}=\frac{2x+1}{\sqrt{x^3}-1}-\frac{x+\sqrt{x}+1}{\sqrt{x^3}-1}\)
\(=\frac{2x+1-(x+\sqrt{x})}{\sqrt{x^3}-1}=\frac{x-\sqrt{x}}{\sqrt{x^3}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(P-\frac{1}{3}=\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}=\frac{2\sqrt{x}-(x+1)}{3(x+\sqrt{x}+1)}\)
\(=\frac{-(\sqrt{x}-1)^2}{3(x+\sqrt{x}+1)}\)
Với \(x\neq 1, x\geq 0\Rightarrow -(\sqrt{x}-1)^2< 0; x+\sqrt{x}+1>0\)
Do đó: \(P-\frac{1}{3}< 0\Rightarrow P< \frac{1}{3}\)
a) \(B=\left(\sqrt{x}-\dfrac{2}{1+\sqrt{x}}\right):\left(\dfrac{1}{1-\sqrt{x}}-\dfrac{2\sqrt{x}}{1-x}\right)=\dfrac{\sqrt{x}+x-2}{1+\sqrt{x}}:\left[\dfrac{1+\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{1+\sqrt{x}}:\dfrac{1+\sqrt{x}-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{1+\sqrt{x}}:\dfrac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)=x+\sqrt{x}-2\)b) Ta có P<10\(\Leftrightarrow x+\sqrt{x}-2< 10\Leftrightarrow x+\sqrt{x}-12< 0\Leftrightarrow x-3\sqrt{x}+4\sqrt{x}-12< 0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-3\right)+4\left(\sqrt{x}-3\right)< 0\Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)< 0\left(1\right)\)
Ta có \(\sqrt{x}+4>0\)
Vậy (1)\(\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
Kết hợp với ĐK
Vậy 0\(\le x< 9,x\ne1\) thì P<10
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)
\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)
\(=\dfrac{1}{2\sqrt{2}a}\)
\(=\dfrac{\sqrt{2}}{4a}\)
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)
chịu đấy :v
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)
\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)
\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)
\(=\dfrac{-x+1+x^2}{x-3}\)
d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)
\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)
\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)
\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)
\(=4x-2\sqrt{2}+\sqrt{x^2}\)
\(=4x-2\sqrt{x}+x\)
\(=5x-2\sqrt{2}\)
a/ \(A=\left(\dfrac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right)\cdot\left(\dfrac{x-1}{2x+\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\right)\cdot\left(\dfrac{x-1}{2x+\sqrt{x}-1}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\cdot\dfrac{x-1}{2x+\sqrt{x}-1}=\dfrac{-3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(2x+2\sqrt{x}-\sqrt{x}-1\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}=\dfrac{-3}{2\sqrt{x}-1}\)
b/ \(A< 0\Leftrightarrow\dfrac{-3}{2\sqrt{x}-1}< 0\)
Ta thấy -3 < 0 nên để A < 0 thì:
\(2\sqrt{x}-1>0\)
\(\Leftrightarrow2\sqrt{x}>1\)
\(\Leftrightarrow\sqrt{x}>\dfrac{1}{2}\Leftrightarrow x>\dfrac{1}{4}\)
Vậy \(x>\dfrac{1}{4}\) thì A < 0
Lời giải:
\(B=\frac{3}{x-1}\sqrt{\frac{(x-1)^2}{(3x)^2}}=\frac{3}{x-1}|\frac{x-1}{3x}|\)
\(=\frac{3}{x-1}.\frac{1-x}{3x}=\frac{-1}{x}\)
\(B=\dfrac{3}{x-1}.\sqrt{\dfrac{x^2-2x+1}{9x^2}}=\dfrac{3}{x-1}.\sqrt{\left(\dfrac{x-1}{3x}\right)^2}\)
\(=\dfrac{3}{x-1}.\left|\dfrac{x-1}{3x}\right|=\dfrac{3}{x-1}.\dfrac{1-x}{3x}=-\dfrac{1}{x}\)