Tính:

a,  . +  : 

 7 . 12 + 16 : 8 

   = 84 + 2

    = 86

b, 72 :  - 

 2.6.3-15

= -13

Lời giải:
a)

\(\sqrt{144}.\sqrt{\frac{49}{69}}\sqrt{0,01}=12.\frac{7}{\sqrt{69}}.0,1=\frac{8,4}{\sqrt{69}}=\frac{42\sqrt{69}}{345}\)

b)

\(\sqrt{0,25}-\sqrt{225}+\sqrt{2,25}=\sqrt{0,5^2}-\sqrt{15^2}+\sqrt{1,5^2}\)

\(=0,5-15+1,5=-13\)

c)

\(72:\sqrt{3^3+3^2}-3\sqrt{5^2-3^2}\)

\(=\frac{72}{\sqrt{36}}-3\sqrt{16}=\frac{72}{6}-3.4=12-12=0\)

\(a,\sqrt{64}-\sqrt{16}+\sqrt{\left(-5\right)^2}\)

\(=8+4+5\)

\(=17\)

\(b,\sqrt{49}+\sqrt{4}-\sqrt{9}.\sqrt{144}\)

\(=7+2-3.12\)

\(=9-36\)

\(=-27\)

\(a;\sqrt{64}-\sqrt{16}+\sqrt{\left(-5\right)^2}\)

\(=8-4+5\)

\(=9\)

\(b;\sqrt{49}+\sqrt{4}-\sqrt{9}.\sqrt{144}\)

\(=7+2-3.12\)

\(=7+2-36\)

\(a)\) \(A=\sqrt{49}-2\sqrt{36}+3\sqrt{4}\)

\(A=7-2.6+3.2\)

\(A=7-12+6\)

\(A=1\)

\(b)\) \(B=\frac{1}{2}\sqrt{\frac{144}{225}}-7\sqrt{100}+4\sqrt{\frac{361}{400}}\)

\(B=\frac{1}{2}.\frac{4}{5}-7.10+4.\frac{19}{20}\)

\(B=\frac{2}{5}-70+\frac{19}{5}\)

\(B=\frac{-329}{5}\)

Chúc bạn học tốt ~ 

a: \(\sqrt{169}-\sqrt{225}\)

\(=\sqrt{13^2}-\sqrt{15^2}\)

=13-15

=-2

b: \(\dfrac{\sqrt{144}}{9}\)

\(=\dfrac{\sqrt{12^2}}{9}\)

\(=\dfrac{12}{9}=\dfrac{4}{3}\)

c: \(\sqrt{18}:\sqrt{2}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)

a,\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{5}\)

\(=|^{ }_{ }2-\sqrt{5}|^{ }_{ }-\sqrt{5}\)

\(=\sqrt{5}-2-\sqrt{5}\)(vì \(2-\sqrt{5}< 0\))

=-2

b,\(\sqrt{16}\cdot\sqrt{25}+\sqrt{256}\cdot\sqrt{64}\)

\(=4\cdot5-16\cdot8=20+128=148\)

c,\(\sqrt{\left(\sqrt{2}-3\right)^2}-\sqrt{\left(5-\sqrt{2}\right)^2}\)

\(=|^{ }_{ }\sqrt{2}-3|^{ }_{ }-|^{ }_{ }5-\sqrt{2}|^{ }_{ }\)

\(=3-\sqrt{2}-5+\sqrt{2}\)(vì \(\sqrt{2}-3< 0;5-\sqrt{2}>0\))

\(=-2\)

cảm ơn

à bạn ơi , mình không hiểu câu c gì à , bạn làm ơn viết lại mình xem, 

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

d. ĐKXĐ: $x>\frac{-2}{3}$

PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \frac{1}{3x+2}=1$

$\Leftrightarrow 3x+2=1$

$\Leftrightarrow x=-\frac{1}{3}$

\(3\sqrt{144}-5\sqrt{49}+\dfrac{1}{2}\sqrt{36}\)

\(=3.12-5.7+\dfrac{1}{2}.6\)

\(=36-35+3=4\)