( x + y )³ - ( x - y )³

= [ ( x + y ) - ( x - y ) ][ ( x + y )² + ( x + y )( x - y ) + ( x - y )² ]

= ( x + y - x + y )( x² + 2xy + y² + x² - y² + x² - 2xy + y² )

= 2y( 3x² + y² )

Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)

\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)

\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)

\(=2y\left(3x^2+y^2\right)\)

Phân tích đa thức thành nhân tử?

Ta có: \(1+6x-6x^2-x^3\)

\(=\left(1-x\right)+\left(7x-7x^2\right)+\left(x^2-x^3\right)\)

\(=\left(1-x\right)+7x\left(1-x\right)+x^2\left(1-x\right)\)

\(=\left(1-x\right)\left(1+7x+x^2\right)\)

1 + 6x - 6x² - x³

= ( 6x - 6x² ) + ( 1 - x³ )

= 6x( 1 - x ) + ( 1 - x )( 1 + x + x² )

= ( 1 - x )[ 6x + ( 1 + x + x² ) ]

= ( 1 - x )( 6x + 1 + x + x² )

= ( 1 - x )( x² + 7x + 1 )

Phân tích đa thức thành nhân tử à?

1) \(\left(x+y\right)^3-x^3-y^3\)

\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)

\(=3xy\left(x+y\right)\)

2) \(x^3+1-x^2-x\)

\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)

\(=\left(x+1\right)\left[x^2-x+1-x\right]\)

\(=\left(x+1\right)\left(x^2-2x+1\right)\)

\(=\left(x+1\right)\left(x-1\right)^2\)

( x + y )³ - x³ - y³

= ( x + y )³ - ( x³ + y³ )

= ( x + y )³ - ( x + y )( x² - xy + y² )

= ( x + y )[ ( x + y )² - ( x² - xy + y² ) ]

= ( x + y )( x² + 2xy + y² - x² + xy - y² )

= 3xy( x + y )

x³ + 1 - x² - x

= ( x³ + 1 ) - ( x² + x )

= ( x + 1 )( x² - x + 1 ) - x( x + 1 )

= ( x + 1 )( x² - x + 1 - x )

= ( x + 1 )( x² - 2x + 1 )

= ( x + 1 )( x - 1 )²

Xét hiệu:

\(a^2+b^2+c^2+d^2+e^2-a\left(b+c+d+e\right)\)

\(=a^2+b^2+c^2+d^2+e^2-ab-ac-ad-ae\)

\(=\left(\frac{a^2}{4}-ab+b^2\right)+\left(\frac{a^2}{4}-ac+c^2\right)+\left(\frac{a^2}{4}-ad+d^2\right)+\left(\frac{a^2}{4}-ae+e^2\right)\)

\(=\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\left(\frac{a}{2}-e\right)^2\)

Do \(\left(\frac{a}{2}-b\right)^2\ge0\forall a,b;\left(\frac{a}{2}-c\right)^2\ge0\forall a,c\);\(\left(\frac{a}{2}-d\right)^2\ge0\forall a,d;\left(\frac{a}{2}-e\right)^2\ge0\forall a,e\)Do đó:

\(\left(\frac{a}{2}-b\right)^2+\left(\frac{a}{2}-c\right)^2+\left(\frac{a}{2}-d\right)^2+\left(\frac{a}{2}-e\right)^2\ge0\)

\(\Rightarrow a^2+b^2+c^2+d^2+e^2-a\left(b+c+d+e\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+e^2\ge a\left(b+c+d+e\right)\)

Dấu"="xảy ra khi \(b=c=d=e=\frac{a}{2}\)

ô kê :))

a² + b² + c² + d² + e² ≥ a( b + c + d + e )

<=> a² + b² + c² + d² + e² ≥ ab + ac + ad + ae

Nhân 4 vào từng vế ta được

<=> 4( a² + b² + c² + d² + e² ) ≥ 4( ab + ac + ad + ae )

<=> 4a² + 4b² + 4c² + 4d² + 4e² ≥ 4ab + 4ac + 4ad + 4ae

<=> 4a² + 4b² + 4c² + 4d² + 4e² - 4ab - 4ac - 4ad - 4ae ≥ 0

<=> ( a² - 4ab + 4b² ) + ( a² - 4ac + 4c² ) + ( a² - 4ad + 4d² ) + ( a² - 4ae + 4e² ) ≥ 0

<=> ( a - 2b )² + ( a - 2c )² + ( a - 2d )² + ( a - 2e )² ≥ 0 ( đúng )

Vậy bđt được chứng minh

Dấu "=" xảy ra <=> b = c = d = e = a/2

Vi a + b + c = 1 nên bt tương đương với \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

Ta có : \(P=abc\left(a+b+c\right)\left(a^2+b^2+c^2\right)\le\frac{1}{3}\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\)( 1 ) 

Mặt khác :\(\left(ab+bc+ca\right)^2\left(a^2+b^2+c^2\right)\le\left(\frac{\left(a+b+c\right)^2}{3}\right)^3=\frac{1}{27}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow P\le\frac{1}{3}.\frac{1}{27}=\frac{1}{81}\)

Dấu "=" xảy ra <=> a = b = c = 1/3

Vậy maxP = 1/81 <=> a = b = c = 1/3

tam giác abc cân tại a 

? câu hỏi là j thế bạn 

Từ giả thiết suy ra: \(a\left(b-c\right)=c\left(a-b\right)\left(1\right)\)

Ta có: \(\frac{1}{c}+\frac{1}{a-b}=\frac{a-b+c}{c\left(a-b\right)}\left(2\right)\)

\(\frac{1}{b-c}-\frac{1}{a}=\frac{a-b+c}{a\left(b-c\right)}\left(3\right)\)

Từ 1,2,3 => đpcm