Lời giải:

$\tan B=\frac{CD}{CB}$

$\tan A=\frac{CD}{AC}$

$\Rightarrow CD=CB.\tan B=AC.\tan C$

$\Leftrightarrow CB.\tan 50^0=AC.\tan 40^0=(CB+1)\tan 40^0$

$\Rightarrow CB(\tan 50^0-\tan 40^0)=\tan 40^0$

$\Rightarrow CB=\frac{\tan 40^0}{\tan 50^0-\tan 40^0}=2,38$ (cm)

$CD=CB\tan B=2,38.\tan 50^0=2,84$ (cm)

Trong tam giác vuông BCD:

\(cot50^0=\dfrac{BC}{CD}\) (1)

Trong tam giác vuông ACD:

\(cot40^0=\dfrac{AC}{CD}=\dfrac{BC+1}{CD}\) (2)

Trừ vế (2) cho (1):

\(\dfrac{BC+1}{CD}-\dfrac{BC}{CD}=cot40^0-cot50^0\)

\(\Rightarrow\dfrac{1}{CD}=cot40^0-cot50^0\)

\(\Rightarrow CD=\dfrac{1}{cot40^0-cot50^0}\approx2,84\left(km\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}x>7\\y>-6\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-7}}=u>0\\\dfrac{1}{\sqrt{y+6}}=v>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}7u-4v=\dfrac{5}{3}\\5u+3v=\dfrac{13}{6}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}21u-12v=5\\20u+12v=\dfrac{26}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{79}{123}\\v=\dfrac{29}{41}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-7}=\dfrac{123}{79}\\\sqrt{y+6}=\dfrac{41}{29}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(\dfrac{123}{79}\right)^2+7=...\\y=\left(\dfrac{41}{29}\right)^2-6=...\end{matrix}\right.\)

ĐKXĐ: x > 7; y > -6

Đặt a= \(\dfrac{1}{\sqrt{x-7}}\); b= \(\dfrac{1}{\sqrt{y+6}}\)
ta được pt: \(\left\{{}\begin{matrix}7a-4b=\dfrac{5}{3}\\5a+3b=2\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}21a-12b=5\\20a+12b=\dfrac{26}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}41a=\dfrac{41}{3}\\7a-4b=\dfrac{5}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-7}}=\dfrac{1}{3}\\\dfrac{1}{\sqrt{y+6}}=\dfrac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-7=9\\y+6=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=30\end{matrix}\right.\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=30\end{matrix}\right.\)

ĐKXĐ: \(x\ge-3;y\ge-1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x+3}=u\ge0\\\sqrt{y+1}=v\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u-2v=2\\2u+v=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2v+2\\2\left(2v+2\right)+v=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=2v+2\\5v=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=2\\v=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{y+1}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

Đầu kiện: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\\sqrt{x}-\sqrt{y}=4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\3\sqrt{x}-3\sqrt{y}=13,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x}+2\sqrt{y}=6\\3\sqrt{x}-3\sqrt{x}+2\sqrt{y}-(-3\sqrt{y})=6-13,5\end{matrix}\right.\)

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

Vì \(5\sqrt{y}\ge0\Rightarrow5\sqrt{y}=-7,5< 0\Rightarrow5\sqrt{y}=-7,5\)vô nghiệm.

Vậy hệ phương trình đã cho vô nghiệm 

ĐKXĐ: \(x\ge0;y\ge0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y}=-7,5\\\sqrt{x}=\sqrt{y}+4,5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=-1,5< 0\left(ktm\right)\\\sqrt{x}=\sqrt{y}+4,5\end{matrix}\right.\)

Vậy hệ đã cho vô nghiệm

ĐKXĐ: \(x\ge0;y\ge0\)

\(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\2\sqrt{x}+3\left(3\sqrt{x}-5\right)=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\11\sqrt{x}=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}=3\sqrt{x}-5\\\sqrt{x}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}10y^2=2\\x^2=3y^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{1}{5}\\x^2=\dfrac{8}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{1}{\sqrt{5}}\\y=\pm\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2-6x^2+9y^2-14y^2=108-74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\-5y^2=34\end{matrix}\right.\)

Vì \(-5y^2=34\Rightarrow y^2=\dfrac{34}{-5}< 0\) vô nghiệm

Vậy hệ phương trình đã cho vô nghiệm

\(\left\{{}\begin{matrix}2x^2+3y^2=36\\3x^2+7y^2=37\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\6x^2+14y^2=74\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\5y^2=-34\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x^2+9y^2=108\\y^2=-\dfrac{34}{5}< 0\left(vô-lý\right)\end{matrix}\right.\)

Vậy hệ pt vô nghiệm

\(\left\{{}\begin{matrix}7x^2+13y=-39\\5x^2-11y=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}77x^2+143y=-429\\65x^2-143y=429\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}143x^2=0\\65x^2-143y=429\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-3\end{matrix}\right.\)