1.a) 4.(12+88) = 4.100=400

b)7²- 6 .[2.8:8 ]=7²-6.[16:8]=7²-6 .2= 49-12=37  ( tui ko chắc lắm )

2.a) x=26-11=15 

b) (4x-15)³=5³ 

4x-15=5 

4x= 5+15

4x=20

vậy x = 20

Bài 2: 

a: Ta có: x+11=26

nên x=15

b: Ta có: \(\left(4x-15\right)^3=125\)

\(\Leftrightarrow4x-15=5\)

hay x=5

b: =100x1000=100000

a)=\(\left[72.\left(-2\right)\right].\left[-50.\left(-2\right)\right]\)

=-144.100

=-14400

b)=(125.8).(-5).(-20)

=1000.100

=100000

c)=\(\left[125.\left(-4\right)\right]\).(-3).(-32)

=-500.96

=-48000

d)=-492.125

=-60500

\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)

\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)

\(=2018\times0=0\)

B-a:tương tự bài A

b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{9}\times1=\frac{7}{9}\)\

thanks

B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)

\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)

thanks

A) 2018*3/4+2018*1/4-2018

=2018*(3/4+1/4)-2018

=2018*4/4-2018

=2018*1-2018

=2018-2018

=0

bn thông cảm mình có việc bận,chút nữa mình trả lời câu B

khó nhìn lắm bn ak

sao pn ko cho 

\(\frac{11}{125}-\frac{17}{18}-\frac{5}{8}+\frac{4}{9}+\frac{17}{14}.\)

thì có phải dễ nhìn hơn ko

a, \(A=\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)

\(=\frac{11}{125}+\left(\frac{-17}{18}+\frac{4}{9}\right)+\left(\frac{-5}{7}+\frac{17}{14}\right)\)

\(=\frac{11}{125}+\frac{-1}{2}+\frac{1}{2}\)

\(=\frac{11}{125}\)

b, \(B=1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)

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

\(=4-3=1\)

c, \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-49}{50}\)

1.

a/ 125³ : 9³ = \(\left(\frac{125}{9}\right)^3\) = \(\frac{1953125}{729}\)

b/ 32⁴. 4³ = \(\left(2^5\right)^4.\left(2^2\right)^3\) = \(2^{20}.2^6\) = \(2^{26}\) =67108864

2. \(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)

=> \(\left(x+\frac{1}{2}\right)^{ }=\frac{2^2}{5^2}\)

=>\(\left(x+\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)

=> \(x+\frac{1}{2}=\frac{2}{5}\)

=> \(x=\frac{2}{5}-\frac{1}{2}=\frac{-1}{10}\)

Vậy \(x=\frac{-1}{10}\)

(-5).8.(-2).(-125)

=[(-5).(-2)].[8.(-125)]

=10.(-1000)

=-10000

b: \(=128\left(-25-89+89-125\right)=128\cdot\left(-150\right)=-19200\)