\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2\)

\(=x^2-3^2-x^2+6x-3^2\)

\(=-9-9+6x\)

\(=6x-18\)

1 ) 

\(\left(x-3\right)\left(x+3\right)-\left(x-3\right)^2\)

\(=x^2-3^2-\left[x^2-3x-3x+9\right]\)

\(=x^2-9-x^2+6x-9\)

\(=6x-18\)

2 ) 

\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(6x+1\right)\left(6x-1\right)\)

\(=\left(6x\right)^2+2.6x.1+1+\left(6x\right)^2-2.6x.1+1-2\left[\left(6x\right)^2-1^2\right]\)

\(=36x^2+12x+1+36x^2-12x+1-2\left(36x^2-1\right)\)

\(=72x^2+1+1-72x^2+2\)

\(=4\)

Gọi tứ giác là ABCD,O là giao điểm của 2 đường chéo

Xét t/g AOB có: OA+OB>AB

Xét t/g BOC có: OB+OC>BC

Xét t/g COD có: OC+OD>CD

Xét t/g AOD có: OA+OD>DA

Do đó: OA+OB+OB+OC+OC+OD+OD+OA>AB+BC+CD+DA

=>2(OA+OB+OC+OD)>AB+BC+CD+DA

=>AC+BD > \(\frac{AB+BC+CD+DA}{2}\) (1)

Xét t/g ABC có: AB+BC > AC

Xét t/g BDC có: BC+DC > BD

Xét t/g CDA có: CD+AD > AC

Xét t/g DAB có: DA+AB > BD

Do đó AB+BC+BC+CD+CD+AD+DA+AB > AC+BD+AC+BD

=>2(AB+BC+CD+DA) > 2(AC+BD)

=>AB+BC+CD+DA > AC+BD (2)

Từ (1) và (2) => đpcm

a)251001

b)8281

c)998001

d)7921

e)249999

f)8096

g)8299

h)3192.42

Gọi 3 số tự nhiên liên tiếp là : \(x;x+1;x+2\left(x\in N\right)\)

Theo bài ra ta có : 

\(\left(x+1\right)\left(x+2\right)-x\left(x+1\right)=140\)

\(\Rightarrow x^2+x+2x+2-x^2-x=140\)

\(\Rightarrow2x+2=140\)

\(\Rightarrow2\left(x+1\right)=140\)

\(\Rightarrow x+1=70\)

\(\Rightarrow x=69\)

\(\Rightarrow\hept{\begin{cases}x+1=70\\x+2=71\end{cases}}\)

Vậy 3 số cần tìm là : 69 ; 70 ; 71

a)2x(2x-y)+2y(x-2y)=\(4x^2-2xy+2xy-4y^2=4x^2-4y^2.\)

b)\(x\left(x^{n-1}+y^{n-1}\right)-y^{n-1}\left(x-y\right)\)=\(x^n+y^n-y^n+y^n=x^n+y^n\)

a ) 

\(5x\left(4x-5\right)-4x\left(5x-6\right)=30\)

\(\Rightarrow20x^2-25x-20x^2+24x=30\)

\(\Rightarrow-x=30\)

\(\Rightarrow x=-30\)

Vậy ...

b ) 

\(2x\left(6-3x\right)+3x\left(2x-5\right)=12\)

\(\Rightarrow12x-6x^2+6x^2-15x=12\)

\(\Rightarrow-3x=12\)

\(\Rightarrow x=-4\)

Vậy ...

a) \(5x\left(4x-5\right)-4x\left(5x-6\right)-30\)

\(\Rightarrow20x^2-25x-20x^2+24x=30\)

\(\Rightarrow-1x=30\)

\(\Rightarrow x=-30\)

Vậy x = -30

b) \(2x\left(6-3x\right)+3x\left(2x-5\right)=12\)

\(\Rightarrow12x-6x^2+6x^2-15x=12\)

\(\Rightarrow-3x=12\)

\(\Rightarrow x=-4\)

Vậy x = -4

\(x^4.\frac{2x^2-x-1}{3}\)

đúng k bn

đúng rồi bạn Thanh Ngân ơi

Giups mình nhé

a)f(x)+g(x)=\(x^5-4x^4-2x^2-7-2x^5+6x^4-2x^2+6.\)

=\(-x^5+2x^4-4x^2-1\)

f(x)-g(x)=\(x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

=\(3x^5-10x^4-13\)

b)f(x)+g(x)=\(5x^4+7x^3-6x^2+3x-7-4x^4+2x^3-5x^2+4x+5\)

=\(x^4+9x^3-11x^2+7x-2\)

f(x)-g(x)=\(5x^4+7x^3-6x^2+3x-7+4x^4-2x^3+5x^2-4x-5\)

=\(9x^4+5x^3-x^2-x-12\)

a ) 

\(f\left(x\right)+g\left(x\right)=x^5-4x^4-2x^2-7+-2x^5+6x^4-2x^2+6\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=\left(x^5-2x^5\right)+\left(6x^4-4x^4\right)-\left(2x^2+2x^2\right)+\left(6-7\right)\)

\(\Rightarrow f\left(x\right)+g\left(x\right)=-x^5+2x^4-4x^2-1\)

\(f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7-\left(-2x^5+6x^4-2x^2+6\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=x^5-4x^4-2x^2-7+2x^5-6x^4+2x^2-6\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=\left(x^5+2x^5\right)-\left(4x^4+6x^4\right)+\left(2x^2-2x^2\right)-\left(6+7\right)\)

\(\Rightarrow f\left(x\right)-g\left(x\right)=3x^5-10x^4-13\)