Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=m\Rightarrow a=bm;c=dm\)

Ta có : \(\dfrac{a.b}{c.d}=\dfrac{b.m.b}{d.m.d}=\dfrac{b^2.m}{d^2.m}=\dfrac{b^2}{d^2}\)(1)

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bm+b\right)^2}{\left(dm+d\right)^2}=\dfrac{\left[b.\left(m+1\right)\right]^2}{\left[d.\left(m+1\right)\right]^2}=\dfrac{b^2.\left(m+1\right)^2}{d^2.\left(m+1\right)^2}=\dfrac{b^2}{d^2}\)(2)

Từ (1) và (2) suy ra :\(\dfrac{a.b}{c.d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Vậy \(\dfrac{a.b}{c.d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) khi \(\dfrac{a}{b}=\dfrac{c}{d}\)

Đc chưa bạn . Tick cho mk nha!

a ) Ta có : \(15^3-25^2=\left(5.3\right)^3-\left(5^2\right)^2=5^3.3^3-5^{2.2}\)

\(=5^3.27-5^4=5^3.27-5^3.5=5^3.\left(27-5\right)=5^3.22\)

Vì \(22⋮11\Rightarrow5^3.22⋮11\Rightarrow\left(15^3-25^2\right)⋮11\)

Vậy \(\left(15^3-25^2\right)⋮11\)

b ) Ta có : \(2^{17}+2^{14}=2^{14}.2^3+2^{14}=2^{14}.8+2^{14}=2^{14}.\left(8+1\right)=2^{14}.9^{14}\)

nhầm

số 9^14 là số 9 nha bạn

nó là : \(2^{14}.\left(8+1\right)=2^{14}.9\)

Vì \(9⋮9\Rightarrow2^{14}.9⋮9\)\(\Rightarrow\)\(\left(2^{17}+2^{14}\right)⋮9\)

Vậy \(\left(2^{17}+2^{14}\right)⋮9\)

Ta có : E= \(1+2^1+2^4+2^7+2^{10}+....+2^{2017}\)

\(\Rightarrow\) \(2^3\) . E =\(2^3.\)( \(1+2^1+2^4+2^7+2^{10}+....+2^{2017}\))

\(\Rightarrow8E=\)\(2^3+2^4+2^7+2^{10}+2^{13}+....2^{2017}+2^{2020}\)

\(\Rightarrow8E-E=\)\(2^3+2^4+2^7+2^{10}+2^{13}+....2^{2017}+2^{2020}\)\(-\)

(\(1+2^1+2^4+2^7+2^{10}+....+2^{2017}\))

\(\Rightarrow\)7E=\(2^3+2^{2020}-\left(1+2^1\right)\)

\(\Rightarrow7E=2^3+2^{2020}-3\)

\(\Rightarrow7E=5+2^{2020}\)

\(\Rightarrow E=\left(5+2^{2020}\right):7\)

Chỉ viết về dạng đó thui bạn nhé :))

Bài làm

Ta có : A = \(^{2017^2}\)=2017.2017=2017.(2016+1)=2017.2016+2017

B = 2016.2018 = 2016.(2017+1)=2016.2017+2016

Vì 2017>2016\(\Rightarrow\) 2017.2016+2017>2016.2017+2016

\(\Rightarrow\) A>B

Vậy A > B

mk chưa hiểu  cân tại A là sao hả bạn

\(^{8^4.16^5.3^2=\left(2^3\right)^4.\left(2^{\text{4}}\right)^5.3^2=2^{12}.2^{20}.3^2=2^{32}.3^2}\)

bn chỉ viết dưới dạng này thui nhé vì tính ra số to lắm!

\(\dfrac{\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{2}{9}-\dfrac{2}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{9}-\dfrac{4}{11}}=\dfrac{2.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}{4.\left[\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right]}\)\(=\dfrac{2}{4}=\dfrac{1}{2}\)