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12 tháng 8 2018

\(\dfrac{x^2-3xy+y^2}{x+y+2}=\dfrac{\left(3+\sqrt{5}\right)^2+3\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\left(3-\sqrt{5}\right)^2}{3+\sqrt{5}+3-\sqrt{5}+2}=\dfrac{9+2\cdot3\sqrt{5}+5+3\left(9-5\right)+9-2\cdot3\cdot\sqrt{5}+5}{8}=\dfrac{40}{8}=5\)

9 tháng 12 2018

5. \(y=\dfrac{-3x}{x+2}\)

xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)

vậy D= (\(-\infty;+\infty\))\{-2}

6. \(y=\sqrt{-2x-3}\)

xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)

vậy D= (\(-\infty;\dfrac{-3}{2}\)]

7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)

xác định khi: x-4 >0 <=> x>4

vậy D= (\(4;+\infty\))

8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)

xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)

vậy D= (\(-\infty;5\))\ {3}

9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)

xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)

vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]

9 tháng 12 2018

1. \(y=\dfrac{3x-2}{x^2-4x+3}\)

xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)

vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)

2.\(y=2\sqrt{5-4x}\)

xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)

vậy D= (\(-\infty;\dfrac{5}{4}\)]

3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)

xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)

vậy D= (\(-3;\dfrac{5}{2}\)]

4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)

xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)

Vậy D= [\(-2;9\)]\{2}

26 tháng 6 2018

b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)

Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)

(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)

\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)

\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)

Kl: x=19, y= 5, z=1890

26 tháng 6 2018

c) \(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) (*)

Đk: \(x\ge5\)

(*) \(\Leftrightarrow3\sqrt{x-5}+x-5-x+14=9+3\sqrt{x-5}\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Vậy nghiệm của phương trình (*) là \(x\ge5\)

a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

17 tháng 8 2018

a) điều kiện xác định : \(x\ge0;x\ne1\)

ta có : \(P=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(\Leftrightarrow P=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(\Leftrightarrow P=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

b) để \(P=\dfrac{2}{3}\Leftrightarrow\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(\sqrt{x}-1\right)=2\left(x+\sqrt{x}+1\right)\Leftrightarrow3\sqrt{x}-3=2x+2\sqrt{x}+2\)

\(\Leftrightarrow2x-\sqrt{x}+5=0\Leftrightarrow2\left(x-\dfrac{1}{2}\sqrt{x}+\dfrac{1}{16}\right)+\dfrac{79}{16}\)

\(\Leftrightarrow2\left(x-\dfrac{1}{4}\right)^2+\dfrac{79}{16}=0\left(vôlí\right)\)

vậy không tồn tại \(x\) để \(P=\dfrac{2}{3}\)

1 tháng 10 2023

a) \(\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{\left(y^2\right)^2}}\) 

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

b) \(\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{\left(x^2y^4\right)^2}}\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{20x^3y^3}{2x^2y^4}\)

\(=\dfrac{10x}{y}\)

c) \(ab^2\sqrt{\dfrac{3}{a^2b^4}}\)

\(=ab^2\dfrac{\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=ab^2\cdot\dfrac{\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

1 tháng 10 2023

\(a,\dfrac{y}{x}\cdot\sqrt{\dfrac{x^2}{y^4}}\left(y\ge0;x,y\ne0\right)\) (sửa đề)

\(=\dfrac{y}{x}\cdot\dfrac{\sqrt{x^2}}{\sqrt{y^4}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{\sqrt{\left(y^2\right)^2}}\)

\(=\dfrac{y}{x}\cdot\dfrac{x}{y^2}\)

\(=\dfrac{1}{y}\)

\(---\)

\(b,\dfrac{5}{2}x^3y^3\cdot\sqrt{\dfrac{16}{x^4y^8}}\left(x,y\ne0\right)\)

\(=\dfrac{5}{2}x^3y^3\cdot\dfrac{\sqrt{16}}{\sqrt{x^4y^8}}\)

\(=\dfrac{5x^3y^3}{2}\cdot\dfrac{4}{x^2y^4}\)

\(=\dfrac{5x\cdot2}{y}\)

\(=\dfrac{10x}{y}\)

\(---\)

\(c,ab^2\sqrt{\dfrac{3}{a^2b^4}}\left(a>0;b\ne0\right)\) (sửa đề)

\(=ab^2\cdot\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\)

\(=\dfrac{ab^2\sqrt{3}}{\sqrt{\left(ab^2\right)^2}}\)

\(=\dfrac{ab^2\sqrt{3}}{ab^2}\)

\(=\sqrt{3}\)

#\(Toru\)

26 tháng 9 2018

\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)

\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)

26 tháng 9 2018

Mk lam sai oy

5 tháng 9 2021

\(a,P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(x\ge0;x\ne9\right)\\ P=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

\(b,x=14-6\sqrt{5}=\left(3-\sqrt{5}\right)^2\)

Thay vào P:

\(P=\dfrac{14-6\sqrt{5}+8}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}=\dfrac{\left(4+\sqrt{5}\right)\left(22-6\sqrt{5}\right)}{11}=\dfrac{55-2\sqrt{5}}{11}\)

 

5 tháng 9 2021

a) \(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(đk:x\ge0,x\ne9\right)\)

\(=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-3x+x\sqrt{x}+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)

b) \(P=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\dfrac{22-6\sqrt{5}}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{3-\sqrt{5}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}\)