Chứng minh đẳng thức sau với a,b,c thuộc Z:

                 a(b-c)-a(b+d)=-a(c+d)

\(ab-ac-ab+ad=-a\left(c+d\right)\)

\(a.\left(b-c-b+d\right)=-a\left(c+d\right)\)

\(-a.\left(c+d\right)\)= VP

\(\Rightarrowđpcm\)

chúc bạn học tốt

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

   \(\Leftrightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)

   \(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng t/c dtsbn:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)

a) a(b + c) - b(a - c)

= ab + ac - ba + bc

= ac + bc

= (a + b)c

b) sorry bạn mình chưa học phần này

a) a ( b + c ) - b ( a - c ) = ab + ac - ab - bc

                                   = [ ab + ( -ab ) ] [ ac + bc ]

                                   = ac + bc

                                   = c ( a + b )

b) Tương tự                             

a, VT= a(b + c) - b(a - c) = ab +ac - ab + bc = ac + bc = c(a+b)

=> VT=VP

=>đpcm

b, Câu b hình như sai đề : mình nghĩ là : a( b - c) - a(b+d)  thì mới đúng .

nếu theo đề mới thì : VT= a( b - c) - a( b + d) = ab -ac -ab -ad = -ac - ad = -a (c+d)

=> VT =VP

=> đpcm 

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\\\dfrac{a}{c}=\dfrac{b}{d}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left(\dfrac{a}{c}\right)^2=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\\left(\dfrac{a}{c}\right)^2=\dfrac{ab}{cd}\end{matrix}\right.\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

a, \(-a-\left(b-a-c\right)=-a-b+a+c=-b+c\)

b, \(-\left(a-c\right)-\left(a-b+c\right)=-a+c-a+b-c=-2a+b\)

c, \(b-\left(b+a-c\right)=b-b-a+c=-a+c\)

d, \(-\left(a-b+c\right)-\left(a+b+c\right)=-a+b-c-a-b-c=-2a-2c\)

Cảm ơn bạn んuﾘ ｲ nhiều lắm!

  a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)

\(a^3+b^3=2\left(c^3-8d^3\right)\)

\(\Leftrightarrow a^3+b^3=2c^3-16d^3\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3c^3-15d^3\)

Ta có: \(3c^3-15d^3=3\left(c^3-5d^3\right)⋮3\)

\(\Rightarrow a^3+b^3+c^3+d^3⋮3\)(1)

Ta có: \(a^3-a=\left(a-1\right)a\left(a+1\right)⋮3\)

\(b^3-b=\left(b-1\right)b\left(b+1\right)⋮3\)

\(c^3-c=\left(c-1\right)c\left(c+1\right)⋮3\)

\(d^3-d=\left(d-1\right)d\left(d+1\right)⋮3\)

\(\Rightarrow a^3+b^3+c^3+d^3-a-b-c-d⋮3\)(2)

Từ (1) và (2) suy ra \(a+b+c+d⋮3\)