giúp mình với:
so sánh
7^39 và 51^20
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Ta có: \(\dfrac{20}{39}>\dfrac{20}{41}>\dfrac{18}{41}\left(1\right)\)
\(\dfrac{22}{27}>\dfrac{22}{29}\left(2\right)\)
Lại có: \(\dfrac{18}{43}=1-\dfrac{25}{43}\)
\(\dfrac{14}{49}=1-\dfrac{25}{39}\)
Vì \(\dfrac{25}{43}< \dfrac{25}{39}\)
⇒ \(1-\dfrac{25}{43}< 1-\dfrac{25}{39}\left(3\right)\)
Từ (1) (2) (3) ⇒ A<B
cai kia sai mk giải lại nha
Ta có \(\dfrac{20}{39}>\dfrac{20}{41}>\dfrac{18}{41}\left(1\right)\)
\(\dfrac{22}{27}>\dfrac{22}{29}\left(2\right)\)
Lại có: \(\dfrac{18}{43}=1-\dfrac{25}{43}\)
\(\dfrac{14}{39}=1-\dfrac{25}{39}\)
Vì: \(\dfrac{25}{43}< \dfrac{25}{39}\)
⇒ \(1-\dfrac{25}{39}>1-\dfrac{25}{43}\left(3\right)\)
Từ (1) (2) (3) ⇒ A>B
9( x + 28 ) = 39 + 51
9( x + 28 ) = 90
x + 28 = 90 : 9
x + 28 = 10
x = -18
\(A=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(B=13\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{6}{43.49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(=\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}{\frac{13}{3}\left(\frac{1}{7}-\frac{1}{49}\right)}=\frac{\frac{17}{3}}{\frac{13}{3}}=\frac{17}{13}\)
A=34/7.13+51/13.22+85/22.37+68/37.49
=17(2/7.13+3/13.22+5/22.37+4/37.49)
=17/3(6/7.13+9/13.22+15/22.37+12/37.49)
=17/3(1/7-1/13+1/13-1/22+1/22-1/37+1/37-1/49)
=17/3(1/7-1/49) b=13(3/7.16+5/16.31+4/31.43+2/43.49)
B=13/3(9/7.16+15/16.31+12/31.43+6/43.49)
=13/3(9/7.16+15/16.31+12/31.43+6/43.49)
=13/3(1/7-1/16+1/16-1/31+1/31-1/43+1/43-1/49)
=13/3(1/7-1/49)
Ta có:
\(7^{39}< 7^{40}=49^{20}\)
\(51^{20}>49^{20}\)
⇒ \(7^{39}< 51^{20}\)