Có \(A=\frac{29^{29}+5}{29^{30}+5}\) và \(B=\frac{29^{30}+5}{29^{31}+5}\)

Xét \(A=\frac{29^{29}+5}{29^{30}+5}\Rightarrow29A=\frac{29^{30}+145}{29^{30}+5}=\frac{29^{30}+5}{29^{30}+5}+\frac{140}{29^{30}+5}=1+\frac{140}{29^{30}+5}\)

Xét \(B=\frac{29^{30}+5}{29^{31}+5}\Rightarrow29B=\frac{29^{31}+145}{29^{31}+5}=\frac{29^{31}+5}{29^{31}+5}+\frac{140}{29^{31}+5}=1+\frac{140}{29^{31}+5}\)

Vì \(1+\frac{140}{29^{30}+5}>1+\frac{140}{29^{31}+5}\Leftrightarrow29A>29B\Leftrightarrow A>B\)

  5/11 . 18/29 - 8/11 . 5/29 + 5/29 . 19/11 

=5/11x(5/29+13/29)-8/11 . 5/29 + 5/29 . 19/11 

=5/11x5/29+13/29x5/11-8/11x5/29+5/29x19/11

=(5/11x5/29+13/29x5/11)-(8/11x5/29-5/29x19/11)

=5/11x(5/29+13/29)-5/29x(8/11-19/11)

=5/11x18/29-5/29x(-1)

=90/319-(-5/29)=5/11

D=5/3.5+5/5.7+5/7.9+5/9.11+...+5/27.29+5/29.31

2D=10/3.5+10/5.7+10/7.9+10/9.11+...+10/27.29+10/29.31

2D=5/3-5/5+5/5-5/7+5/7-5/9+5/9-5/11+...+5/27-5/29+5/29-5/31

2D=5/3-5/31

2D=155/93-15/93

2D=140/93

D=140/93:2

D=70/93

a: \(\dfrac{5}{13}\left(\dfrac{6}{29}-\dfrac{26}{39}\right)-\dfrac{6}{29}\cdot\left(\dfrac{5}{13}-\dfrac{29}{6}\right)\)

\(=\dfrac{5}{13}\cdot\dfrac{6}{29}-\dfrac{5}{13}\cdot\dfrac{26}{39}-\dfrac{6}{13}\cdot\dfrac{5}{13}+\dfrac{6}{29}\cdot\dfrac{29}{6}\)

\(=\dfrac{-5}{39}\cdot2+1=1-\dfrac{10}{39}=\dfrac{29}{39}\)

b: \(\dfrac{1\cdot198+2\cdot197+3\cdot196+...+198\cdot1}{1\cdot2+2\cdot3+...+198\cdot199}\)

\(=\dfrac{1\left(199-1\right)+2\left(199-2\right)+...+198\cdot\left(199-198\right)}{1\left(1+1\right)+2\left(1+2\right)+...+198\left(1+198\right)}\)

\(=\dfrac{199\left(1+2+...+198\right)-\left(1^2+2^2+...+198^2\right)}{\left(1+2+...+198\right)+\left(1^2+2^2+...+198^2\right)}\)

\(=\dfrac{199\cdot\dfrac{198\cdot199}{2}-\dfrac{198\cdot\left(198+1\right)\cdot\left(2\cdot198+1\right)}{6}}{198\cdot\dfrac{199}{2}+\dfrac{198\left(198+1\right)\left(2\cdot198+1\right)}{6}}\)

\(=\dfrac{3\cdot198\cdot199^2-198\cdot199\cdot397}{6}:\dfrac{3\cdot198\cdot199+198\cdot199\cdot397}{6}\)

\(=\dfrac{198\cdot199\left(3\cdot199-397\right)}{198\cdot199\left(3+397\right)}\)

\(=\dfrac{200}{400}=\dfrac{1}{2}\)

pls mai mình phải nộp rồi

7/23.5/29+7/29.18/23-7/29

=(7.5)/(23.29)+7/29.18/23-7/29

=7/29.5/23+7/29.18/23-7/29.1

=7/29.(5/23+18/23-1)

=7/29.(1-1)

=7/29.0

=0

\(\frac{7}{23}\cdot\frac{5}{29}+\frac{7}{29}\cdot\frac{18}{23}-\frac{7}{29}\)

\(=\frac{7}{29}\cdot\frac{5}{23}+\frac{7}{29}\cdot\frac{18}{23}-\frac{7}{29}\)

\(=\frac{7}{29}\cdot\left(\frac{5}{23}+\frac{18}{23}-1\right)\)

\(=\frac{7}{29}\cdot0\)

\(=0\)

\(a,\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\\ b,\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\\ =\sqrt{1}=1\)

a: \(\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\)

b: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)

=1