chịu

a) \(\dfrac{5}{7}-1\dfrac{4}{7}\left(450\%+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{5}{7}-\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{-1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{5}{7}+\dfrac{1}{14}\)

\(\dfrac{11}{7}\left(\dfrac{9}{2}+\dfrac{2}{3}x\right)=\dfrac{11}{14}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{11}{14}:\dfrac{11}{7}=\dfrac{11}{14}.\dfrac{7}{11}\)

\(\dfrac{9}{2}+\dfrac{2}{3}x=\dfrac{1}{2}\)

\(\dfrac{2}{3}x=\dfrac{1}{2}-\dfrac{9}{2}=-4\)

\(x=-4:\dfrac{2}{3}=-4.\dfrac{3}{2}=-6\)

Vậy x = \(-6\)

b) \(100=6.7^{\left|x+2\right|}-194\)

\(100+194=6.7^{\left|x+2\right|}\)

\(294=6.7^{\left|x+2\right|}\)

\(294:6=49=7^{\left|x+2\right|}\)

\(\Rightarrow7^2=7^{\left|x+2\right|}\)

\(\Rightarrow2=\left|x+2\right|\Rightarrow\pm2=x+2\)

+ x + 2 = -2 \(\Rightarrow\) x = - 4

+ x + 2 = 2 \(\Rightarrow\) x = 0

Vậy x = - 4 hoặc 0

\(\dfrac{-270}{450}=\dfrac{-9}{15}\)
\(\dfrac{11}{-143}=\dfrac{1}{-13}\)
\(\dfrac{32}{12}=\dfrac{8}{3}\)
\(\dfrac{-26}{-156}=\dfrac{1}{6}\)

a) 450 – ( 25 – 10) = 450 – 15

                   = 435

450 – 25 – 10 = 425 – 10

                   = 415

b) 180 : 6 : 2 = 30 : 6

                   = 15

180 : ( 6 : 2 ) = 180 : 3

                   = 60

c) 410 – (50 +30) = 410 -80

                   = 330

410 - 50 + 30 = 360 + 30

                   = 390

d) 16 x (6 : 3) = 16 x 2

                   = 32

16 x 6 : 3 = 96 : 3

                   = 32

1:

a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)

b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)

=>2x-1>0

=>x>1/2

2:

a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)

\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)

\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)

\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)

\(=45\sqrt{2}-19\sqrt{5}\)

b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)

\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)

\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)