Ta có \(A=\dfrac{2}{xy}+\dfrac{3}{x^2+y^2}=\dfrac{4}{2xy}+\dfrac{3}{x^2+y^2}=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\)Lại có \(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\ge\dfrac{4}{x^2+2xy+y^2}=\dfrac{4}{\left(x+y\right)^2}=4\) (vì \(x+y=1\))

Và \(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\Leftrightarrow2xy\le\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2xy}\ge2\)

Do đó \(A\ge3.4+2=14\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)

Vậy GTNN của A là 14 khi \(x=y=\dfrac{1}{2}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ge0\\3-2x\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le\dfrac{3}{2}\end{matrix}\right.\)

PT\(\Leftrightarrow x-1+4\left(\sqrt{x+3}-2\right)+2\left(\sqrt{3-2x}-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)+4.\dfrac{x+3-4}{\sqrt{x+3}+2}+2.\dfrac{3-2x-1}{\sqrt{3-2x}+1}=0\)

\(\Leftrightarrow\left(x-1\right)+4.\dfrac{x-1}{\sqrt{x+3}+2}-2.\dfrac{2\left(x-1\right)}{\sqrt{3-2x}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left[1+\dfrac{4}{\sqrt{x+3}+2}-\dfrac{4}{\sqrt{3-2x}+1}\right]=0\)

TH1:\(x-1=0\Leftrightarrow x=1\left(tmđk\right)\)

TH2:\(1+\dfrac{4}{\sqrt{x+3}+2}-\dfrac{4}{\sqrt{3-2x}+1}=0\)(VN)

Đề bài không cho biết phần tô đậm trong hình vẽ chỗ nào!!!

\(\left\{{}\begin{matrix}\dfrac{3}{x-1}+\dfrac{4}{y}=13\\\dfrac{2}{x-1}-\dfrac{5}{y}=1\end{matrix}\right.\)(1)

ĐK: \(\left\{{}\begin{matrix}x-1\ne0\\y\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\y\ne0\end{matrix}\right.\)

Đặt \(u=\dfrac{1}{x-1};v=\dfrac{1}{y}\)

\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}3u+4v=13\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6u+8v=26\\6u-15v=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}23v=23\\2u-5v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}v=1\\2u=1-5v=1+5.1=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=1\\u=\dfrac{6}{2}=3\end{matrix}\right.\)

- Khi u= 3, ta có \(\dfrac{1}{x-1}=3\Leftrightarrow1=3\left(x-1\right)\Leftrightarrow1=3x-3\)

\(\Leftrightarrow3x=4\Leftrightarrow x=\dfrac{4}{3}\)(thỏa mãn)

- Khi v= 1, ta có: \(\dfrac{1}{y}=1\Leftrightarrow y=1\)(thỏa mãn)

Vậy nghiệm của hệ phương trình là: \(\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=1\end{matrix}\right.\)

1) y = x^2/5
Tọa độ A có tung độ bằng 5 -> y = 5
-> x^2 = 25
-> x = 5 hoặc x = -5
2)Do phương trình có 2 nghiệm phân biệt ta có
 x1x2 = c/a = m^2
 x1+x2 = -b/a = 2(m+3)
 (x1+x2)^2 = x1^2 + 2x1x2 +x2^2
 = 15 + x1x2 = 15 + m^2
 -> (2(m+3) )^2 = 15 + m^2
 -> 4(m^2 +6m + 9) = 15 + m^2
 -> 3m^2 +24m + 21 = 0
 delta = 24*24 -4*3*21 =324, sqrt(delta) = 18
 
 m1 = (-24+18)/6 = -1
 m2 = (-24-18)/6 = -7
 Vậy m = -1 hoặc m = -7

1, vt : \(\left(1-\dfrac{5+\sqrt{2}}{\sqrt{2}+1}\right).\sqrt{3+2\sqrt{2}}\)

=\(\dfrac{\sqrt{2}+1-5-\sqrt{2}}{\sqrt{2}+1}.\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1}\)

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

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

=-4

2, A=\(\left(\dfrac{\sqrt{x}}{x+\sqrt{x}}-\dfrac{1}{\sqrt{x}-1}\right)\div\dfrac{2}{x+\sqrt{x}-2}\)

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

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

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

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

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