\(D=x+2y-\sqrt{2x-1}-5\cdot\sqrt{4y-3}+13\)

\(=\frac12\left(2x+4y-2\sqrt{2x-1}-10\sqrt{4y-3}+26\right)\)

\(=\frac12\left\lbrack2x-1-2\sqrt{2x-1}+1+4y-3-10\cdot\sqrt{4y-3}+25+4\right\rbrack\)

\(=\frac12\left\lbrack\left(\sqrt{2x-1}-1\right)^2+\left(\sqrt{4y-3}-5\right)^2+4\right\rbrack\ge\frac12\cdot\left\lbrack0+0+4\right\rbrack=2\forall x,y\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\begin{cases}2x-1=1\\ 4y-3=25\end{cases}\Rightarrow\begin{cases}2x=2\\ 4y=28\end{cases}\Rightarrow\begin{cases}x=1\\ y=7\end{cases}\)

bài đầu tiên bằng -3

bài thứ hai mình ko biết

Dễ =))

Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

\(\Rightarrow A_{min}=\dfrac{5}{2}\)

đúng như mk dự đoán chớ mk thủ hết cách rk mà có  dc à

Có : \(x-2y-\sqrt{xy}+\sqrt{x}-2\sqrt{y}=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{x}-2\sqrt{y}=0\)

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

\(\Leftrightarrow\sqrt{x}=2\sqrt{y}\) (Do \(\sqrt{x}+\sqrt{y}+1>0,\forall x;y>0\))

\(\Leftrightarrow x=4y\)

Khi đó \(P=\dfrac{7y}{\left(2\sqrt{y}+3\sqrt{y}\right).\left(\sqrt{x}+2\sqrt{y}\right)}\)

\(=\dfrac{7y}{5\sqrt{y}.4\sqrt{y}}=\dfrac{7}{20}\)

ĐKXĐ: 5-x>=0 và x+8>=0

=>-8<=x<=5

Ta có: \(13\sqrt{5-x}+18\sqrt{x+8}=61+x+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\sqrt{5-x}-26+18\sqrt{x+8}-54=x-19+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\left(\sqrt{5-x}-2\right)+18\left(\sqrt{x+8}-3\right)=x-1-18+3\sqrt{\left(5-x\right)\left(x+8\right)}\)

=>\(13\cdot\frac{5-x-4}{\sqrt{5-x}+2}+18\cdot\frac{x+8-9}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{\left(5-x\right)\left(x+8\right)}-6\right)\)

=>\(13\cdot\frac{1-x}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{5x+40-x^2-8x}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\left(\sqrt{-x^2-3x+40}-6\right)\)

=>\(-13\cdot\frac{\left(x-1\right)}{\sqrt{5-x}+2}+18\cdot\frac{x-1}{\sqrt{x+8}+3}=x-1+3\cdot\frac{-x^2-3x+40-36}{\sqrt{-x^2-3x+40}+6}\)

=>(x-1)\(\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-3x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{-x^2-4x+x+4}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1+3\cdot\frac{\left(x+4\right)\left(-x+1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}\right)=x-1-3\cdot\frac{\left(x+4\right)\left(x-1\right)}{\sqrt{-x^2-3x+40}+6}\)

=>\(\left(x-1\right)\left(-\frac{13}{\sqrt{5-x}+2}+\frac{18}{\sqrt{x+8}+3}-1+3\cdot\frac{x+4}{\sqrt{-x^2-3x+40}+6}\right)=0\)

=>x-1=0

=>x=1(nhận)

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...

Ta có:

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

Áp dụng bđt Minkowski, ta có:

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

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

\(A=\sqrt{4^2+\left(\sqrt{2}+\sqrt{3}\right)^2\left(y+1\right)^2}\ge\sqrt{4^2}=4\)

\(\Rightarrow A\ge4.Đ\text{TXR}\Leftrightarrow\orbr{\begin{cases}x=1;y=-1\\x=3;y=-1\end{cases}}\)

Dấu "=" xảy ra khi (x; y) = (3; -1)