\(\Leftrightarrow16-x+2\sqrt{\left(16-x\right)\left(9+x\right)}+9+x=49\)

\(\Leftrightarrow2\sqrt{\left(16-x\right)\left(9+x\right)}=24\)

\(\Leftrightarrow\left(16-x\right)\left(9+x\right)=144\)

\(\Leftrightarrow7x-x^2=0\)

\(\Leftrightarrow x\left(7-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=7\end{cases}}\)

đặt ĐK cửa x

Do 2 vế không âm nên bình phương hai vế ta có

25 + 2\(\sqrt{\left(16-x\right)\left(9+x\right)}=49\)

\(\sqrt{\left(16-x\right)\left(9+x\right)}=12\)

Do hai vế không âm nên bình phương hai vế ta có

(16-x)(9+x) = 144

144 + 7x - \(x^2=144\)

\(x^2-7x=0\)

X = 0; 7

ĐK: \(-9\le x\le16\)

\(\sqrt{16-x}=a;\sqrt{9+x}=b\ge0\Rightarrow a^2+b^2=25\)

Kết hợp đề bài ta có hệ:

\(\hept{\begin{cases}a^2+b^2=25\\a+b=7\end{cases}}\Leftrightarrow\hept{\begin{cases}a^2+b^2=25\\\left(a+b\right)^2=49\end{cases}}\)

Nhân chéo hai vế của hai pt cho nhau ta được: \(49\left(a^2+b^2\right)=25\left(a+b\right)^2\Leftrightarrow2\left(3a-4b\right)\left(4a-3b\right)=0\)

...

căn (16+9)=5

và căn 16+ căn 9=4+3=7

\(\sqrt{16}-\sqrt{9}+\sqrt{16+9}-\sqrt{\left(-2\right)^2}\)

\(=4-3+5-2=4\)

\(e,\sqrt{\dfrac{9}{169}}=\dfrac{\sqrt{9}}{\sqrt{169}}=\dfrac{\sqrt{3^2}}{\sqrt{13^2}}=\dfrac{3}{13}\)

\(f,\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{\sqrt{25}}{\sqrt{16}}=\dfrac{\sqrt{5^2}}{\sqrt{4^2}}=\dfrac{5}{4}\)

\(g,\dfrac{\sqrt{2300}}{\sqrt{23}}=\sqrt{\dfrac{2300}{23}}=\sqrt{100}=\sqrt{10^2}=10\)

\(h,\dfrac{\sqrt{12,5}}{\sqrt{0,5}}=\sqrt{\dfrac{12,5}{0,5}}=\sqrt{25}=\sqrt{5^2}=5\)

e, \(\sqrt{\dfrac{9}{169}}\)

\(=\sqrt{\dfrac{3^2}{13^2}}\)

\(=\dfrac{3}{13}\)

f, \(\sqrt{1\dfrac{9}{16}}\)

\(=\sqrt{\dfrac{25}{16}}\)

\(=\sqrt{\dfrac{5^2}{4^2}}\)

\(=\dfrac{5}{4}\)

g, \(\dfrac{\sqrt{2300}}{\sqrt{23}}\)

\(=\dfrac{10\sqrt{23}}{\sqrt{23}}\)

\(=10\)

h, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)

\(=\dfrac{\dfrac{5\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}\)

\(=\dfrac{\dfrac{5\sqrt{2}}{2}\cdot2}{\sqrt{2}}\)

\(=\dfrac{5\sqrt{2}}{\sqrt{2}}=5\)

\(\left(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\right)\)

\(=\left(\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\right)\)

\(=\left(\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}\right)\)

\(=\left(\frac{5}{4}-\frac{3}{4}\right)=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}\)

...Vậy ...................

\(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}=\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\)

\(=\sqrt{\left(\frac{5}{4}\right)^2}-\sqrt{\left(\frac{3}{4}\right)^2}=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\)

#

\(\sqrt{1\frac{9}{16}}-\sqrt{\frac{9}{16}}\)

\(=\sqrt{\frac{25}{16}}-\sqrt{\frac{9}{16}}\)

\(=\frac{5}{4}-\frac{3}{4}=\frac{2}{4}=\frac{1}{2}\)

\(=20-12+5-6\)

\(=7\)

= 20 - 12 + 5 - 6

= 7

\(=\left(\dfrac{3}{2}-3\right).\sqrt{\dfrac{25}{16}}=\left(-\dfrac{3}{2}\right).\dfrac{5}{4}=-\dfrac{15}{8}\)