\(\left[1-\left(\frac{3}{4}-\frac{2}{3}\right)\right]-\left[1-\left(\frac{5}{3}-\frac{1}{4}\right)\right]-\left[1-\left(\frac{4}{3}+\frac{3}{4}\right)\right]\)

\(=\left[1-\frac{3}{4}+\frac{2}{3}\right]-\left[1-\frac{5}{3}+\frac{1}{4}\right]-\left[1-\frac{4}{3}-\frac{3}{4}\right]\)

\(=1-\frac{3}{4}+\frac{2}{3}-1+\frac{5}{3}-\frac{1}{4}-1+\frac{4}{3}+\frac{3}{4}\)

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

\(=-1-\frac{1}{4}+\frac{11}{3}=\frac{29}{12}\)

[3-(8-11)]-[-2+(-15+3)]

=[3+3]-[-2+(-15+3)]

=[3+3]-[-2+-12]

=6-[-2-12]

=6+14

=20

b) Xét \(x=0\)thì \(0+\left|y\right|< 20\)=> \(\left|y\right|< 20\Rightarrow y\in\left\{0;\pm1;\pm2;...;\pm19\right\}\)gồm 39 giá trị

Xét x = \(\pm1\)thì y \(\in\left\{0;\pm1;\pm2;\pm3;...;\pm18\right\}\)gồm 37 giá trị

....

Xét x = \(\pm\)18 thì y \(\in\){0; \(\pm\)1} 

Xét x = \(\pm19\)=> y = 0 , có 1 giá trị

Có tất cả : 2(1 + 3 + ... + 37) + 39 = 761(cặp số)

\(\frac{\left(2^6\right)^2.\left(3^4\right)^3.34}{2^{13}.3^9.17}=\frac{2^{12}.3^{12}.2}{2^{13}.3^9}=3^3=27\)

thanks

\(M=\frac{2x^2+4x+60}{x^2+2x+4}=\frac{2\left(x^2+2x+4\right)+52}{x^2+2x+4}=2+\frac{52}{x^2+2x+4}=2+\frac{52}{\left(x+1\right)^2+3}\)

Để M đạt GTNN => \(\frac{52}{\left(x+1\right)^2+3}\)đạt GTLN

=> \(\left(x+1\right)^2+3\)(*) đạt GTNN

\(\left(x+1\right)^2\ge0\forall x\Rightarrow\left(x+1\right)^2+3\ge3\)

=> Min(*) = 3 <=> x + 1 = 0 => x = -1

=> MinM = \(2+\frac{52}{\left(-1+1\right)^2+3}=2+\frac{52}{3}=\frac{58}{3}\), đạt được khi x = -1

Mình không chắc nha -.-

\(M=\frac{2x^2+4x+60}{x^2+2x+4}=\frac{2\left(x^2+2x+4\right)+52}{x^2+2x+4}=2+\frac{52}{x^2+2x+4}\)

Để M đạt GTLN  => \(\frac{52}{x^2+2x+4}\)(**) đạt GTLN 

Hay \(x^2+2x+4\)(*) đạt GTNN 

Ta có : \(x^2+2x+4=\left(x^2+2x+1\right)+3=\left(x+1\right)^2+3\)

Do \(\left(x+1\right)^2\ge0\forall x\Leftrightarrow\left(x+1\right)^2+3\ge3\forall x\)

Nên GTNN (*) = 3 khi x + 1 = 0 <=> x = -1

Suy ra GTLN (**) = 52/3 khi x = -1

Vậy nên GTLN M = 2 + 52/3 = 58/3 khi x = -1

Ta có : 

-18/173<0

1/2013>0 

Suy ra -18/173<1/2013

thank bạn nha

Ta có :

\(3.9^3.27^2=3.\left(3^2\right)^3.\left(3^3\right)^2\)

\(=3.3^6.3^6=3^{13}\)

3.9^3.27^2

= 3.(3^2)^3.(3^4)^2

=3.3^6.3^8

=3^15