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1
1 tháng 10 2018

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14 tháng 5 2021

a )

`VP= (a+b)^3-3ab(a+b)`

     `=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`

     `=a^3+b^3 =VT (đpcm)`

b) 

b) Ta có

`VT=a3+b3+c3−3abc`

     `=(a+b)3−3ab(a+b)+c3−3abc`

     `=[(a+b)3+c3]−3ab(a+b+c)`

     `=(a+b+c)[(a+b)2+c2−c(a+b)]−3ab(a+b+c)`

     `=(a+b+c)(a2+b2+2ab+c2−ac−bc−3ab)`

    `=(a+b+c)(a2+b2+c2−ab−bc−ca)=VP`

  
14 tháng 5 2021

 

a) Ta có:

`VP= (a+b)^3-3ab(a+b)`

     `=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`

     `=a^3 + b^3=VT(dpcm)`

b) Ta có

`VT=a^3+b^3+c^3−3abc`

     `=(a+b)^3−3ab(a+b)+c^3−3abc`

     `=[(a+b)^3+c^3]−3ab(a+b+c)`

     `=(a+b+c)[(a+b)^2+c^2−c(a+b)]−3ab(a+b+c)`

     `=(a+b+c)(a^2+b^2+2ab+c^2−ac−bc−3ab)`

    `=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)=VP`

d) Ta có: \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

21 tháng 1 2022

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

\(VP=\left(ac+bd\right)^2+\left(ad-bc\right)^2=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2=a^2c^2+b^2c^2+a^2d^2+b^2d^2\)

\(\Rightarrow VT=a^2c^2+b^2c^2+a^2d^2+b^2d^2=VP\left(đpcm\right)\)

b, Tham khảo:Chứng minh hằng đẳng thức:(a+b+c)3= a3 + b3 + c3 + 3(a+b)(b+c)(c+a) - Hoc24

22 tháng 8 2018

a) Áp dụng nhiều lần công thức \(\left(x+y\right)^3=x^3-y^3+3xy\left(x+y\right)\), ta có:

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

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

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(Đpcm\right)\)

b) Ta có:

\(a^3+b^3+c^3-3abc\)

\(=a^3+3ab\left(a+b\right)+b^2+c^3-3abc-3ab\left(a+b\right)\)

\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)

Mình nghĩ bằng thế này mới đúng, bạn chắc ghi sai đề rồi bucminh

22 tháng 8 2018

a) Ta có: (a + b + c)3 - a3 - b3 - c3 = [ (a + b + c)3 - a3 ] - ( b3 + c3)

= (a + b + c - a) ( a2 + b2 + c2 + 2ab + 2bc + 2ac + a2 + ab + ac + a2) - (b + c) ( b2 - bc + c3)

= (b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac) - (b + c) ( b2 - bc + c3)

= ( b + c) ( 3a2 + b2 + c2 + 3ab + 2bc + 3ac - b2 + bc - c3)

= ( b + c) ( 3a2 + 3ab + 3bc + 3ac)

= 3 (b + c) [a (a + b) + c (a + b)]

= 3 (b + c) (a + b) (a + c) (đpcm)

DD
28 tháng 7 2021

a) \(\left(a+b+c\right)^2+a^2+b^2+c^2\)

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

b) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2+bc+c^2\right)\)

\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

27 tháng 11 2023

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)

=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

=>\(2\left(ab+bc+ac\right)=0\)

=>ab+bc+ac=0

\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

=>\(\dfrac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^3}=\dfrac{3}{abc}\)

=>\(\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)

\(\Leftrightarrow\left(ab+bc\right)^3-3\cdot ab\cdot bc\cdot\left(ab+bc\right)+\left(ac\right)^3=3\left(abc\right)^2\)

=>\(\left(-ac\right)^3-3\cdot ab\cdot bc\cdot\left(-ac\right)+\left(ac\right)^3-3\left(abc\right)^2=0\)

=>\(-a^3c^3+a^3c^3+3a^2b^2c^2-3a^2b^2c^2=0\)

=>0=0(đúng)