Trả lời

Phần a bạn chỉ cần đưa mũ về rồi tính

Phần b bạn lậy 2 số giống nhau trừ đi rồi chuyển vế cho nó = 0 rồi tính kết quả 

study well 

chỗ 3^2x4 nghĩa lad 2 nhân 4 à

a) a=9*y

    b=9*x

do đó a+b = 9*y+9*x=72

=9*(y+x)=72

x+y=8

ta có bảng sau

vậy (x,y) thuộc{1,7;7,1;3,5;5,3;4,4;2,6;6,2;}

b) a=14*x

b=14*y

a*b=7840=14*x*14*y

7840/14/14=x*y

x*y=40

ta có bảng sau: tương tự câu a

\(ƯCLN_{\left(51,70\right)}=1\)

Có \(51=3.17;70=2.5.7;\)

Vậy \(UCLN\left(51;70\right)=1\)

#)Giải :

\(\left(1-\frac{1}{15}\right)\left(1-\frac{1}{21}\right)\left(1-\frac{1}{28}\right)...\left(1-\frac{1}{210}\right)=\frac{14}{15}\times\frac{20}{21}\times\frac{27}{28}\times...\times\frac{209}{210}\)

\(=\frac{28}{30}\times\frac{40}{42}\times\frac{54}{56}\times...\times\frac{418}{420}=\frac{4\times7}{5\times6}\times\frac{5\times8}{6\times7}\times\frac{6\times9}{7\times8}\times...\times\frac{19\times22}{20\times21}\)

\(=\frac{4\times5\times6\times...\times19}{5\times6\times7\times...\times20}\times\frac{7\times8\times9\times...\times22}{6\times7\times8\times...\times21}=\frac{4}{20}\times\frac{22}{6}=\frac{11}{15}\)

\(\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right).....\left(1-\frac{1}{210}\right)\)

\(=\left(\frac{15}{15}-\frac{1}{15}\right).\left(\frac{21}{21}-\frac{1}{21}\right).\left(\frac{28}{28}-\frac{1}{28}\right).....\left(\frac{210}{210}-\frac{1}{210}\right)\)

\(=\frac{14}{15}.\frac{20}{21}.\frac{27}{28}....\frac{209}{210}\)

\(=\frac{2.7}{3.5}.\frac{5.4}{7.3}.\frac{3.9}{4.7}....\frac{11.19}{21.10}\)

\(=\frac{2}{3}.\frac{19}{10}\)

\(=\frac{19}{15}\)

GIẢI GIÚP MK NHÉ!

\(5^{36}=\left(5^3\right)^{12}=125^{12}\)

\(11^{24}=\left(11^2\right)^{12}=121^{12}\)

\(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

Ta có: \(5^{36}=\left(5^3\right)^{12}=125^{12}\) 

^{\(11^{24}=\left(11^2\right)^{12}=121^{12}\)} 

Vì 121¹² < 125¹²

Nên 11²⁴ < 5³⁶

Ta có : A = 1 + 2 + 3 + ... + 2008

\(A=\frac{\left(2008+1\right)\left[\left(2008-1\right)\div1+1\right]}{2}\) 

\(A=\frac{2009.2008}{2}\) 

\(A=2017036\) 

Ta có: B = 1 + 2 + 3 + ... + 1010

\(B=\frac{\left(1010+1\right)\left[\left(1010-1\right):1+1\right]}{2}\) 

\(B=\frac{1011.1010}{2}\) 

\(B=510555\)

\(A=1+2+3+4+5+...+2008\)

\(A=\left(2008+1\right)\left(\left(2008-1\right):1+1\right):2=2009.2008:2\)

\(=2009.1004=2017036\)

\(B=1+2+3+4+...+1010\)

\(B=\left(1010+1\right)\left(\left(1010-1\right):1+1\right):2=1011.\left(1010:2\right)\)

\(=1011.505=510555\)

\(C=2+5+8+11+...+302\)

\(C=\left(302+2\right)\left(\left(302-2\right):3+1\right):2=304.101:2\)

\(=15352\)

\(D=3+3^2+3^3+3^4+...+3^{2019}\)

\(3D=3^2+3^3+3^4+...+3^{2020}\)

\(3D-D=\left(3^2+3^3+3^4+...+3^{2020}\right)-\left(3+3^2+3^3+3^4+...+3^{2019}\right)\)

\(2D=3^{2020}-3\)

\(\Rightarrow D=\frac{3^{2020}-3}{2}\)

\(E=4^{10}+4^{11}+4^{12}+...+4^{100}\)

\(4E=4^{11}+4^{12}+4^{13}+...+4^{101}\)

\(4E-E=\left(4^{11}+4^{12}+4^{13}+...+4^{101}\right)-\left(4^{10}+4^{11}+4^{12}+...+4^{100}\right)\)

\(3E=4^{101}-4^{10}\)

\(E=\frac{4^{101}-4^{10}}{3}\)