Câu 1:

a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}\)

b,Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}\)

\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ac}{cd}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)

Ta lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}\Rightarrow\frac{ab}{cd}=\left(\frac{a+b}{c+d}\right)^2\left(2\right)\)

Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)

Câu 2:

\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+....+a2018}\)

\(\Rightarrow\frac{a1}{a2}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(1\right)\)

\(\frac{a2}{a3}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2\right)\)

..............

\(\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2017\right)\)

Nhân các vế (1),(2)....(2017) ta được:

\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\cdot\cdot\cdot\cdot\frac{a2017}{a2018}=\frac{a1}{a2018}=\left(\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\right)^{2017}\)

Vậy...

Câu 3:

\(x_2^2=x_1x_3\Rightarrow\frac{x1}{x2}=\frac{x2}{x3}\)

\(x_3^2=x_2x_4\Rightarrow\frac{x2}{x3}=\frac{x3}{x4}\)

\(x_4^2=x_3x_5\Rightarrow\frac{x3}{x4}=\frac{x4}{x5}\)

\(x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}\)

Đến đây thfi làm giống câu 2

cho x₁, x₂ , x₃ là 3 số thực khác 0 thỏa mãn x₁ + x₂ + x₃ = a ; x₁x₂ + x₂x₃ + x₁x₃ = 0 ; x₁x₂x₃ = b

CMR: a/b < 0

a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)

\(=142-\left[50-80+40\right]\)

=142-10

=132

a. 142 - [50-(2³.10-2³.5)]

= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]

= 142 - [ 50 - ( 80 - 40 )]

= 142 - [ 50 - 40 ]

= 142 - 10 = 132

b. 375 : { 32[ 4+(5.3² - 42 )]} - 14

= 375 : { 32[4+(5.9 - 42 )]} - 14

= 375 : { 32[4 + ( 45 - 42 )]} - 14

= 375 : {32[4+3]} - 14

= 375 : 224 - 14

c.{210 : [ 16+3.(6+3.2^2)]} - 3

= {210 : [ 16 + 3.(6+3.4)]} - 3

= { 210 :[16+3.(6+12)]}-3

= {210 : [ 16+3.18)]} - 3

= { 210 : [ 16 + 54]} - 3

= { 210 : 70 } - 3

= 30 - 3 = 27

d.500-{5.[409-(2^3 . 3-21^2)] - 1724}

= 500-{5.[409-(8 . 63 - 21^2] - 1724}

= 500 - { 5.[409- (504 -  441) - 1724}

= 500 -{ 5.[ 409 - 63 ] - 1724}

= 500 - { 5.346 - 1724}

= 500 - { 1720 - 1724 }

= 500 - 6 

= 494

a,= (x+4)\(^3\)

b,= (x-2)\(^3\)

c,= x\(^3\)+8

d,=x\(^3\)-27

a,= x\(^3\)+3.4x\(^2\)+3.4\(^2\).x+4\(^3\)=(x+4)\(^3\)

-3.2.x\(^2\)+3.2\(^2\).x-2\(^3\)= (x-2)\(^3\)

còn c,d áp dụng HĐT là ra! ( đc chx bà nội)

\(a,\frac{a+b}{a-b}=\frac{c+a}{c-a}\Rightarrow\frac{a+b}{c+a}=\frac{a-b}{c-a}=\frac{a+b+a-b}{c+a+c-a}=\frac{2a}{2c}=\frac{a}{c}\)

\(\text{Suy ra: }\frac{a+b}{c+a}=\frac{a}{c}\Rightarrow c.\left(a+b\right)=a.\left(c+a\right)\Rightarrow ac+bc=ac+a^2\)

=>a²=bc

b)Viết đề rõ lại giúp