Có :

\(x=\dfrac{1}{\sqrt{5}-2}\Rightarrow x^2=\dfrac{1}{\left(\sqrt{5}-2\right)^2}=\dfrac{1}{5-4\sqrt{5}+4}\\ =\dfrac{1}{9-4\sqrt{5}}\\ y=\dfrac{1}{5+4\sqrt{5}}=\dfrac{1}{5+4\sqrt{5}+2}=\dfrac{1}{\left(\sqrt{5}+2\right)^2}\\ \Rightarrow\sqrt{y}=\sqrt{\dfrac{1}{\left(\sqrt{5}+2\right)^2}}=\dfrac{1}{\sqrt{5}+2}\) 

\(\Rightarrow A=\dfrac{1}{9-4\sqrt{5}}-3.\dfrac{1}{\sqrt{5}-2}.\dfrac{1}{\sqrt{5}+2}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{1}{9-4\sqrt{5}}-\dfrac{3}{5-4}+\dfrac{2}{9+4\sqrt{5}}\\ =\dfrac{9+\sqrt{5}+2\left(9-4\sqrt{5}\right)}{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}-3=\dfrac{27-4\sqrt{5}}{81-80-3}\\ =27-4\sqrt{5}-3=24-4\sqrt{5}\)

Đúng là "trẩu" chị anh hùng bàn phím là nhanh :^

1) Ta có: \(\left\{{}\begin{matrix}2x+y=5\\3x-2y=11\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x+3y=15\\6x-4y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=-7\\2x+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\2x=5-y=5-\left(-1\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

2) Ta có: \(B=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+3\sqrt{x}+2+2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{1}{\sqrt{x}+2}\)

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

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

\(=3\sqrt{x}\)

5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)

Thay từng TH rồi làm nha bạn

3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)

thay nhá

Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)

PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)

+) Với y = x - 1 thay vào pt (2):

\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))

Anh quy đồng lên đê, chắc cần vài con trâu đó:))

+) Với y = 2x + 3...

1. \(y'=\sqrt{x-2}+\dfrac{x+1}{2\sqrt{x-2}}\)

2. \(y'=-\dfrac{\dfrac{1}{2\sqrt{x^2+4x+5}}\cdot\left(x^2+4x+5\right)'}{x^2+4x+5}=-\dfrac{x+2}{\sqrt{\left(x^2+4x+5\right)^3}}\)

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

4. \(y'=\dfrac{\sqrt{x^2+1}-\dfrac{x+1}{2\sqrt{x^2+1}}\cdot\left(x^2+1\right)'}{x^2+1}=\dfrac{\dfrac{2\left(x^2+1\right)-\left(x+1\right)\cdot2x}{2\sqrt{x^2+1}}}{x^2+1}=\dfrac{1-x}{\sqrt{\left(x^2+1\right)^3}}\)

5. \(y'=-\dfrac{\dfrac{\left(4-3x^2\right)'}{2\sqrt{4-3x^2}}}{4-3x^2}=\dfrac{3x}{\sqrt{\left(4-3x^2\right)^3}}\)

1. \(y'=\sqrt{x-2}+\dfrac{x+1}{2\sqrt{x-2}}=\dfrac{3x-3}{2\sqrt{x-2}}\)

2. \(y'=-\dfrac{\left(\sqrt{x^2+4x+5}\right)'}{x^2+4x+5}=-\dfrac{x+2}{\left(x^2+4x+5\right)\sqrt{x^2+4x+5}}\)

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

4. \(y'=\dfrac{\sqrt{x^2+1}-\dfrac{x\left(x+1\right)}{\sqrt{x^2+1}}}{x^2+1}=\dfrac{1-x}{\left(x^2+1\right)\sqrt{x^2+1}}\)

5. \(y'=\dfrac{\left(\sqrt{4-3x^2}\right)'}{3x^2-4}=\dfrac{-3x}{\left(3x^2-4\right)\sqrt{4-3x^2}}\)

mọi người giúp mk với

câu 6 sai nha

sửa : \(\dfrac{1}{x}+\dfrac{3}{2y+1}=2\)

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

hỏi trước tí, bạn biết giải cái hệ này chứ?

\(\left\{{}\begin{matrix}2x+y=3\\2x-3y=1\end{matrix}\right.\)

VAG

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

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}