\(A=\left(37,1-4,5\right)-\left(-4,5+37,1\right)\)

   \(=37,1-4,5+4,5-37,1\)

   \(=37,1-37,1-4,5+4,5\)

   \(=0.\)

\(B=-\left(315.4+275\right)+4.315-\left(10-275\right)\)

   \(=-315.4-275+4.315-10+275\)

   \(=-315.5+4.315-275+275-10\)

   \(=0+0-10=-10.\)

\(C=-\left(\frac{3}{7}+\frac{3}{8}\right)-\left(-\frac{3}{8}+\frac{4}{7}\right)\)

    \(=-\frac{3}{7}-\frac{3}{8}+\frac{3}{8}-\frac{4}{7}\)

    \(=-\frac{3}{7}-\frac{4}{7}-\frac{3}{8}+\frac{3}{8}\)

    \(=-1+0=-1.\)

A=37,1 - 4,5 + 4,5 - 37,1

A=(37,1 - 37,10) + ( 4,5 - 4,5 )

 A = 0 + 0 = 0

B= -315.4 - 275  + 4.315 - 10 + 275

B=(-315.4 + 4.315) + ( 275-275) - 10

B= 0 + 0 - 10 = -10

C= -3/7 - 3/8 + 3/8 - 4/7

C = ( -3/7-4/7) + ( 3/8 - 3/8)

C=-7/7 + 0 = -7/7 = -1

f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)

a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)

b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)

c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)

a: 2x(x-1/7)=0

=>x(x-1/7)=0

=>x=0 hoặc x=1/7

b: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

nên \(x=\dfrac{-1}{4}:\dfrac{7}{20}=\dfrac{-20}{4\cdot7}=\dfrac{-5}{7}\)

c: \(\Leftrightarrow\dfrac{41}{9}:\dfrac{41}{18}-7< x< \left(3.2:3.2+\dfrac{45}{10}\cdot\dfrac{31}{45}\right):\left(-21.5\right)\)

\(\Leftrightarrow2-7< x< \dfrac{\left(1+3.1\right)}{-21.5}\)

\(\Leftrightarrow-5< x< \dfrac{-41}{215}\)

mà x là số nguyên

nên \(x\in\left\{-4;-3;-2;-1\right\}\)

Bn k có máy tính ạ/

Bn bẫm mày thui kém thế Nên tự làm đi

a. \(\dfrac{3}{4}-\left|2x+1\right|=\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}\)

=> \(\left|2x+1\right|=\dfrac{-1}{8}\)

=> \(\left\{{}\begin{matrix}2x+1=\dfrac{-1}{8}\\2x+1=\dfrac{1}{8}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{-9}{16}\\x=\dfrac{-7}{16}\end{matrix}\right.\)

#Yiin

b. \(2.\left|2x-3\right|=\dfrac{1}{2}\)

=> \(\left|2x-3\right|=\dfrac{1}{4}\)

=> \(\left\{{}\begin{matrix}2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)

a. \(\dfrac{-39}{7}:x=26\)

x = \(\dfrac{-39}{7}:26\)

x = \(\dfrac{-3}{14}\)

b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)

x = \(\dfrac{7}{4}.\dfrac{13}{5}\)

x = \(\dfrac{91}{20}\)

c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)

x = \(\dfrac{-11}{10}\)

d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)

x = \(\dfrac{9}{4}+\dfrac{3}{4}\)

x = 3

e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)

x = \(\dfrac{7}{8}:\dfrac{14}{3}\)

x = \(\dfrac{3}{16}\)

f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)

x = \(\dfrac{13}{3}.\dfrac{8}{3}\)

x = \(\dfrac{104}{9}\)

g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)

x = 0

chúc bạn học tốt

a, \(A=\dfrac{10^{15}+1}{10^6+1}>1\);\(B=\dfrac{10^6+1}{10^{17}+1}< 1\)

⇒\(A>B\)

b, \(D=\dfrac{2^{2007}+3}{2^{2006}-1}=\dfrac{2^{2008}+6}{2^{2007}-2}\)

Ta có : \(\dfrac{2^{2008}-3}{2^{2007}-1}< \dfrac{2^{2008}-3}{2^{2007}-2}< \dfrac{2^{2008}+6}{2^{2007}-2}\)

⇒ \(C< D\)

c, \(M=\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)

\(N=\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)

Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\)

⇒ \(M< N\)

8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)

=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)

=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)

=7.(-7)

=-49