Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)

Thanks nha!!!

ồ, lâu h ms gặp

a,

Dễ thấy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)

Áp dụng khi \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\left(n\in N^{\circledast}\right)\)

Ta có:

\(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2016}+1+\left(2005^2-1\right)}{2005^{2017}+1+\left(2005^2-1\right)}=\dfrac{2005^{2016}+2005^2}{2005^{2017}+2005^2}=\dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)

Vậy \(\dfrac{2005^{2016}+1}{2005^{2017}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)

b,

\(\dfrac{19}{10}=\dfrac{10+9}{10}=\dfrac{10}{10}+\dfrac{9}{10}=1+\dfrac{9}{10}\\ \dfrac{49}{40}=\dfrac{40+9}{40}=\dfrac{40}{40}+\dfrac{9}{40}=1+\dfrac{9}{40}\)

Vì \(10< 40\Rightarrow\dfrac{9}{10}>\dfrac{9}{40}\Rightarrow1+\dfrac{9}{10}>1+\dfrac{9}{40}\Leftrightarrow\dfrac{19}{10}>\dfrac{49}{40}\)Vậy \(\dfrac{19}{10}>\dfrac{49}{40}\)

c,

\(\dfrac{13}{20}=\dfrac{20-7}{20}=\dfrac{20}{20}-\dfrac{7}{20}=1-\dfrac{7}{20}\\ \dfrac{33}{40}=\dfrac{40-7}{40}=\dfrac{40}{40}-\dfrac{7}{40}=1-\dfrac{7}{40}\)

Vì \(20< 40\Rightarrow\dfrac{7}{20}>\dfrac{7}{40}\Rightarrow1-\dfrac{7}{20}< 1-\dfrac{7}{40}\Leftrightarrow\dfrac{13}{20}< \dfrac{33}{40}\)

Vậy \(\dfrac{13}{20}< \dfrac{33}{40}\)

Áp dụng tính chất:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

\(\)Đặt: \(B=\dfrac{2005^{2016}+1}{2005^{2017}+1}< 1\)

\(\Rightarrow B< \dfrac{2005^{2016}+1+4020024}{2005^{2017}+1+4020024}\)

\(B< \dfrac{2005^{2016}+4020025}{2005^{2017}+4020025}\)

\(B< \dfrac{2005^2\left(2005^{2014}+1\right)}{2005^2\left(2005^{2015}+1\right)}\)

\(B< \dfrac{2005^{2014}+1}{2005^{2015}+1}=A\)

\(B< A\)

cho hỏi chút

\(\frac{a}{b}=\frac{c}{d}\)

trong đó

\(a=c\) hay \(a\ne c\)

\(b=d\) hay \(b\ne d\)

( bài có thiếu điều kiện ko vậy )

câu B là \(2^{12}\) nha mấy bn

mk chỉ giải đc câu d thôi nha ; bn thông cảm

d) \(D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}\)

\(D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1\)

giả sử a/2002 = b/2003 = c/2004 = k
=> a = 2002k ; b=2003k và c=2004k
=> 4(a-b)(b-c) = 4(2002k - 2003k)(2003k - 2004k)
=> 2(a-b)(b-c) = 4k^2 (1)
Ta có (c-a)^2 = (2004k - 2002k)^2 = 4k^2 (2)
từ (1) và (2) ta có 2(a-b)(b-c) = (c-a)^2

mk ko hiểu chỗ từ dòng số 3 đến dòng số 4 cho lắm .

Giảng cho mk dc ko ?

Đặt \(\dfrac{a}{2003}=\dfrac{b}{2005}=\dfrac{c}{2007}=k\Rightarrow\left\{{}\begin{matrix}a=2003k\\b=2005k\\c=2007k\end{matrix}\right.\)

Ta có: \(\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(2003k-2007k\right)^2}{4}=\dfrac{16k^2}{4}=4k^2\) (1)

\(\left(a-b\right)\left(b-c\right)=\left(2003k-2005k\right)\left(2005k-2007k\right)\)

\(=2k2k=4k^2\) (2)

Từ (1), (2) \(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\left(a-b\right)\left(b-c\right)\left(đpcm\right)\)

Vậy...

a: \(A=3^n\cdot27+5^n\cdot125+3^n\cdot3+5^n\cdot25\)

\(=3^n\cdot30+5^n\cdot150\)

Vì \(3^n\cdot30\) chia 60 dư 30(do 3ⁿ là số lẻ)

và \(5^n\cdot150\) chia 60 dư 30(do 5ⁿ là số lẻ)

nên A chia hết cho 60

c: a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a-b}{c-d}\right)^{2003}=\left(\dfrac{bk-b}{dk-d}\right)^{2003}=\left(\dfrac{b-1}{d-1}\right)^{2003}\)

\(\dfrac{a^{2005}+b^{2005}}{c^{2005}+d^{2005}}=\dfrac{b^{2005}k^{2005}+b^{2005}}{d^{2005}k^{2005}+d^{2005}}=\dfrac{b^{2005}}{d^{2005}}\)

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