B =
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. a) \(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\
=\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\
=\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\) b) \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\
=\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\
=\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\
=\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\) c) \(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\) d) \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\
=\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\
=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\
=\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\
=x-\sqrt{x}-x-\sqrt{x}+x+1\\
=x-2\sqrt{x}+1\\
=\left(x-1\right)^2\) \(\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}+\frac{\sqrt{2}+1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}+2}-\frac{\sqrt{2}}{\left(1+\sqrt{2}\right)\sqrt{2}}+\frac{\left(\sqrt{2}+1\right)^2}{\sqrt{2}\left(\sqrt{2}+1\right)}=\frac{\sqrt{2}-1}{2+\sqrt{2}}-\frac{\sqrt{2}}{2+\sqrt{2}}+\frac{3+2\sqrt{2}}{2+\sqrt{2}}=\frac{\sqrt{2}-1-\sqrt{2}+3+2\sqrt{2}}{2+\sqrt{2}}=\frac{2+2\sqrt{2}}{2+\sqrt{2}}\) \(b,\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}=\left(\sqrt{x}-2\right)+\frac{10-x}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+10-x}{\sqrt{x}+2}=\frac{x-4+10-x}{\sqrt{x}+2}=\frac{6}{\sqrt{x}+2}\) \(c,\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\) a: \(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\) \(=\dfrac{2}{x+\sqrt{x}+1}\) b: \(A-2=\dfrac{2-2x-2\sqrt{x}-2}{x+\sqrt{x}+1}\) \(=\dfrac{-2x-2\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{-2\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}< =0\) =>A<=2 Vì \(x+\sqrt{x}+1>0\) nên A>0 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: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\) \(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\) \(1.\) Gỉa sử : \(\sqrt{25-16}< \sqrt{25}-\sqrt{16}\) \(\Leftrightarrow3< 1\) ( Vô lý ) \(\Rightarrow\sqrt{25-16}>\sqrt{25}-\sqrt{16}\) \(2.\sqrt{a}-\sqrt{b}< \sqrt{a-b}\) \(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2< a-b\) \(\Leftrightarrow a-2\sqrt{ab}+b< a-b\) \(\Leftrightarrow2b-2\sqrt{ab}< 0\) \(\Leftrightarrow2\left(b-\sqrt{ab}\right)< 0\) Ta có :\(a>b\Leftrightarrow ab>b^2\Leftrightarrow\sqrt{ab}>b\) \(\RightarrowĐpcm.\) \(2a.\) Áp dụng BĐT Cauchy , ta có : \(a+b\ge2\sqrt{ab}\left(a;b\ge0\right)\) \(\Leftrightarrow\dfrac{a+b}{2}\ge\sqrt{ab}\) \(b.\) Áp dụng BĐT Cauchy cho các số dương , ta có : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\left(x,y>0\right)\left(1\right)\) \(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(y,z>0\right)\left(2\right)\) \(\dfrac{1}{x}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{xz}}\left(x,z>0\right)\left(3\right)\) Cộng từng vế của ( 1 ; 2 ; 3 ) , ta được : \(2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\) \(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\) \(3a.\sqrt{x-4}=a\left(a\in R\right)\left(x\ge4;a\ge0\right)\) \(\Leftrightarrow x-4=a^2\) \(\Leftrightarrow x=a^2+4\left(TM\right)\) \(3b.\sqrt{x+4}=x+2\left(x\ge-2\right)\) \(\Leftrightarrow x+4=x^2+4x+4\) \(\Leftrightarrow x^2+3x=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\) KL.... b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\) c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\) \(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\) e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\) F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\) \(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\) \(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\) \(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\) \(\Rightarrow\sqrt{x}+3\) \(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\) \(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\) \(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\) \(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\) \(\Rightarrow\sqrt{y}-1\) \(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) \(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\) \(\Rightarrow\sqrt{xy}\) \(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\) \(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\) \(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\) \(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\) \(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\) \(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\) \(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\) \(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\) Bài 1: a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\) \(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\) \(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\) b: \(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}\) \(=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{1}\cdot\dfrac{\sqrt{x}-1}{2}\) \(=-\sqrt{x}\left(\sqrt{x}-1\right)\) a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\) \(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\) b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\) \(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\) \(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\) g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\) \(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\) \(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
K
Khách
NH
Nguyễn Huy Tú
21 tháng 7 2021
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