b, 2.|x+y|.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 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) \(\sqrt{\dfrac{9x^2}{25}}+\dfrac{1}{5}x\) (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. a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\) b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) \(\sqrt{x}=a,\sqrt{y}=b\) Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\) \(\Rightarrow B=x+\sqrt{xy}+y\) Vậy... c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\) d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\) a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4) = \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\) =\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\) = 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn) b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0) =\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\) =\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\) =\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức ) = (x+\(\sqrt{xy}\)+y) c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2) Tương tự câu a d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3) tương tự câu a e:2x +√1−6x+9x23x−1 = \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\) = 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức) =2x+\(\dfrac{3x-1}{3x-1}\) =2x+1 a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\) b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\) \(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\) c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\) \(=\sqrt{x}+2-\sqrt{x}-2=0\) a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\) \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\) \(=4+\sqrt{11}-3\sqrt{7}\) b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\) \(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\) \(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\) b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\) c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\) d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\) e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\) 1: \(\left\{{}\begin{matrix}\left|x-1\right|+\dfrac{2}{y}=2\\-\left|x-1\right|+\dfrac{4}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{6}{y}=3\\\left|x-1\right|=2-\dfrac{2}{y}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=2-\dfrac{2}{2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{2;0\right\}\end{matrix}\right.\) 2: \(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{5}{y-1}=-3\\2\left|x-1\right|+\dfrac{4}{y-1}=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{9}{y-1}=-9\\\left|x-1\right|+\dfrac{2}{y-1}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=2\\\left|x-1\right|=3-\dfrac{2}{2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x\in\left\{3;-1\right\}\end{matrix}\right.\) 3: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x-5}+\dfrac{12}{\sqrt{y}-2}=4\\\dfrac{2}{x-5}-\dfrac{1}{\sqrt{y}-2}=-9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{13}{\sqrt{y}-2}=13\\\dfrac{1}{x-5}=2-\dfrac{6}{\sqrt{y}-2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=9\\\dfrac{1}{x-5}=2-\dfrac{6}{3-2}=2-\dfrac{6}{1}=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=9\\x-5=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{19}{4}\\y=9\end{matrix}\right.\)
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Nguyễn Lê Phước Thịnh
11 tháng 8 2023
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