XIN LỖI Ơ PHẦN B=1+3+3^2+...+3^8

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                                                              Bài giải

\(A=\frac{1+5+5^2+5^3+...+5^9}{1+5+5^2+...+5^8}=1+\frac{5^9}{1+5+5^2+...+5^8}\)

Đặt \(C=1+5+5^2+..+5^8\)

\(5C=5+5^2+5^3+...+5^9\)

\(5C-C=4C=5^9-1\)

\(C=\frac{5^9-1}{4}\)

Thay vào ta được : \(A=\frac{5^9}{\frac{5^9-1}{4}}=1+\frac{5^9}{4\cdot5^9-4}=1+\frac{5^9}{4\left(5^9-1\right)}=1+\frac{5^9-1}{4\left(5^9-1\right)}+\frac{1}{4\left(5^9-1\right)}\)

\(=1+\frac{1}{4}+\frac{1}{4\left(5^9-1\right)}=\frac{5}{4}+\frac{1}{4\left(5^9-1\right)}\)

\(B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}=1+\frac{3^9}{1+3+3^2+...+3^8}\)

Đặt \(D=1+3+3^2+...+3^8\)

\(3D=3+3^2+3^3+...+3^9\)

\(3D-D=2D=3^9-1\)

\(D=\frac{3^9-1}{2}\)

Thay vào ta được : \(B=1+\frac{3^9}{\frac{3^9-1}{2}}=1+\frac{3^9}{2\cdot3^9-2}=1+\frac{3^9}{2\left(3^9-1\right)}=1+\frac{3^9-1}{2\left(3^9-1\right)}+\frac{1}{2\left(3^9-1\right)}\)

\(=1+\frac{1}{2}+\frac{1}{2\left(3^9-1\right)}=\frac{3}{2}+\frac{1}{2\left(3^9-1\right)}\)

Vì \(\frac{5}{4}< \frac{3}{2}\) và \(\frac{1}{4\left(5^9-1\right)}< \frac{1}{2\left(3^9-1\right)}\) \(\Rightarrow\text{ }A< B\)

ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)

mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)

\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)

\(\Rightarrow A< B\)

Câu hỏi của nguyen van nam - Toán lớp 6 - Học toán với OnlineMath

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

kieu nay la ko tinh ra ket qua hay so sanh

A=1+C; voi C=5^9/(1+...5^8)=1/(1/5^9+1/5^8+...+1/5)

B=1+D;voi D=3^9/(1+..3^8)=1/(1/3^9+1/3^8+...+1/3)

C=1/E; voi E=(1/5^9+1/5^8+...+1/5)

D=1/f; voi F=(1/3^9+1/3^8+...+1/3)

=> F-E=(1/3-1/5)+...+(1/3^9-1/5^9) >0=> F>E

=> C>D=> A>B