\(=-5\cdot4+0.5-3\cdot\dfrac{4}{5}=-19.5-\dfrac{12}{5}=-\dfrac{219}{10}\)

\(=\dfrac{1}{2}.1,1-0,3+6=6,25\)

\(=\left(\dfrac{3}{2}\cdot\dfrac{2}{5}+2\cdot\dfrac{1}{5}\right):\dfrac{3}{8}=\left(\dfrac{3}{5}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}=\dfrac{8}{3}\)

a: \(\left(\dfrac{5}{9}-\dfrac{\sqrt{9}}{12}\right):\dfrac{3}{4}+\dfrac{11}{3}:\dfrac{3}{4}\)

\(=\left(\dfrac{5}{9}-\dfrac{3}{12}\right)\cdot\dfrac{4}{3}+\dfrac{11}{3}\cdot\dfrac{4}{3}\)

\(=\left(\dfrac{5}{9}-\dfrac{1}{4}+\dfrac{11}{3}\right)\cdot\dfrac{4}{3}\)

\(=\dfrac{20-9+132}{36}\cdot\dfrac{4}{3}\)

\(=\dfrac{143}{3}\cdot\dfrac{1}{9}=\dfrac{143}{27}\)

b: \(\left(0.\left(3\right)+\dfrac{\left|-2\right|}{3}\right):\dfrac{\sqrt{25}}{4}-\left(2^3+3^2\right)^0\)

\(=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\cdot\dfrac{4}{5}-1\)

\(=\dfrac{4}{5}-1=-\dfrac{1}{5}\)

\(=4\cdot5-2\cdot\dfrac{2}{3}=20-\dfrac{4}{3}=\dfrac{56}{3}\)

a:

ĐKXĐ: x>=0; x<>1

Sửa đề: \(P=\left(\frac{x-1}{x+3\sqrt{x}-4}+\frac{\sqrt{x}+1}{1-\sqrt{x}}\right):\frac{x+2\sqrt{x}+1}{x-1}+1\)

\(=\left(\frac{x-1}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right):\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+1\)

\(=\frac{x-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}-1}{\sqrt{x}+1}+1\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}+1\right)}+1=\frac{-5}{\sqrt{x}+4}+1=\frac{-5+\sqrt{x}+4}{\sqrt{x}+4}=\frac{\sqrt{x}-1}{\sqrt{x}+4}\)

b: P<0

=>\(\sqrt{x}-1<0\)

=>\(\sqrt{x}<1\)

=>0<=x<1

Lời giải:
ĐKXĐ: $x\geq 0$

$\sqrt{x}=\frac{5}{\sqrt{x}+2}$

$\Rightarrow \sqrt{x}(\sqrt{x}+2)=5$

$\Rightarrow x+2\sqrt{x}-5=0$

$\Leftrightarrow (\sqrt{x}+1)^2-6=0$

$\Leftrightarrow (\sqrt{x}+1-\sqrt{6})(\sqrt{x}+1+\sqrt{6})=0$

$\Leftrightarrow \sqrt{x}+1-\sqrt{6}=0$ (do $\sqrt{x}+1+\sqrt{6}>0$)

$\Leftrightarrow \sqrt{x}=\sqrt{6}-1$

$\Leftrightarrow x=7-2\sqrt{6}$ (tm)

a) Ta có: \(3\sqrt{2}+4\sqrt{8}-\sqrt{18}\)

\(=\sqrt{2}\left(3+4\cdot2-3\right)\)

\(=8\sqrt{2}\)

b) Ta có: \(\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\)

\(=\sqrt{3}\left(1-\frac{1}{3}\cdot\sqrt{9}+2\cdot\sqrt{169}\right)\)

\(=\sqrt{3}\left(1-1+26\right)\)

\(=26\sqrt{3}\)

c) Ta có: \(\sqrt{25a}+\sqrt{49a}-\sqrt{64a}\)

\(=\sqrt{25}\cdot\sqrt{a}+\sqrt{49}\cdot\sqrt{a}-\sqrt{64}\cdot\sqrt{a}\)

\(=\sqrt{a}\left(5+7-8\right)\)

\(=4\sqrt{a}\)

d) Ta có: \(-\sqrt{36b}-\frac{1}{3}\sqrt{54b}+\frac{1}{5}\sqrt{150b}\)

\(=-\sqrt{6b}\cdot\sqrt{6}-\frac{1}{3}\cdot\sqrt{6b}\cdot\sqrt{9}+\frac{1}{5}\cdot\sqrt{6b}\cdot\sqrt{25}\)

\(=-\sqrt{6b}\left(\sqrt{6}+1-1\right)\)

\(=-\sqrt{6b}\cdot\sqrt{6}=-6\sqrt{b}\)