Gọi hai số nguyên liên tiếp đó là \(n;n-1\left(n\inℤ\right)\)

\(\Rightarrow n^2-\left(n-1\right)^2=n^2-\left(n^2-2n+1\right)=2n-1\)

Vì \(\hept{\begin{cases}2n⋮2\\1⋮̸2\end{cases}\Rightarrow2n-1⋮̸}2\)

\(\Rightarrow2n-1\)là số lẻ

\(\Rightarrowđpcm\)

Gọi 2 số nguyên liên tiếp lần lượt là x và x + 1

Theo bài:

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

Vì 2x là số chẵn => 2x + 1 là số lẻ ( đpcm )

(x + 2)² - (x + 3)(x - 3) = 5

<=> x² + 4x + 4 - x² + 9 = 5

<=> 4x = -8

<=> x = -2

Trả lời:

( x + 2 )² - ( x + 3 ) ( x - 3 ) = 5

<=> x² + 4x + 4 - ( x² - 9 ) = 5

<=> x² + 4x + 4 - x² + 9 = 5

<=> 4x + 13 = 5

<=> 4x = - 8

<=> x = - 2

Vậy x = - 2 là nghiệm của pt.

Trả lời:

a, \(A=\frac{x^2-9}{x^2-6x+9}=\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)^2}=\frac{x+3}{x-3}\)

b, \(B=\frac{9x^2-16}{3x^2-4x}=\frac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\frac{3x+4}{x}\)

c, \(C=\frac{x^2+4x+4}{2x+4}=\frac{\left(x+2\right)^2}{2\left(x+2\right)}=\frac{x+2}{2}\)

d, \(D=\frac{2x-x^2}{x^2-4}=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=-\frac{x}{x+2}\)

e, \(E=\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)

combo tập trung hiệu quả=))

kết bn đi nèo

\(x^3-y^3=xy+25\)

\(\Leftrightarrow\left(x-y\right)^3+3xy\left(x-y\right)=xy+25\)

\(\Leftrightarrow a^3-25=b\left(1-3a\right)\)(\(a=x-y,b=xy\))

\(\Rightarrow a^3-25⋮\left(1-3a\right)\)

\(\Rightarrow27\left(a^3-25\right)⋮\left(3a-1\right)\)

\(\Leftrightarrow27a^3-1-674=\left(3a-1\right)\left(9a^2+3a+1\right)-674⋮\left(3a-1\right)\)

suy ra \(3a-1\inƯ\left(674\right)=\left\{-674,-337,-2,-1,1,2,337,674\right\}\)

Suy r a\(a\in\left\{-112,0,1,225\right\}\).

suy ra các cặp \(\left(a,b\right)\)thỏa mãn là \(\left(-112,-4169\right),\left(0,-25\right),\left(1,12\right),\left(225,-16900\right)\)

suy ra cặp \(\left(x,y\right)\)thỏa mãn là: \(\left(4,3\right),\left(-3,-4\right)\). 

Trả lời:

\(M=\left(x-2020\right)^4+\left(x+y+1\right)^2+5\)

Ta có: \(\left(x-2020\right)^4\ge0\forall x;\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2\ge0\forall x,y\)

\(\Rightarrow\left(x-2020\right)^4+\left(x+y+1\right)^2+5\ge5\forall x,y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2020=0\\x+y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2020\\y=-2021\end{cases}}}\)

Vậy GTNN của M = 5 khi x = 2020; y = - 2021

Trả lời:

Ta có: ( a + b )³ - 3ab . ( a + b ) = a³ + 3a²b + 3ab² + b³ - 3a²b - 3ab² = a³ + b³    (đpcm)

Ta có: ( a - b )³ + 3ab . ( a - b ) = a³ - 3a²b + 3ab² - b³ + 3a²b - 3ab² = a³ - b³      (đpcm)

Trả lời:

a, ( 2x + y )³ - ( 2x - y )³ 

= ( 2x )³ + 3.( 2x )².y + 3.2x.y² + y³ - [ ( 2x )³ - 3.( 2x )².y + 3.2x.y² - y³ ] 

= 8x³ + 12x²y + 6xy² + y³ - ( 8x³ - 12x²y + 6xy² - y³ )

= 8x³ + 12x²y + 6xy² + y³ - 8x³ + 12x²y - 6xy² + y³

= 24x²y + 2y³

b, ( 5 - 3x )³ - ( 5 + 3x )³ 

= 5³ - 3.5².3x + 3.5.( 3x )² - ( 3x )³ - [ 5³ + 3.5².3x + 3.5.( 3x )² + ( 3x )³ ]

= 125 - 225x + 135x² - 27x³ - ( 125 + 225x + 135x² + 27x³ )

= 125 - 225x + 135x² - 27x³ - 125 - 225x - 135x² - 27x³

=  - 54x³- 450x​

c, ( x + 3 ) ( x² - 2x + 9 ) - ( 54 + x³ )

= x³ - 2x² + 9x + 3x² - 6x + 27 - 54 - x³

= x² + 3x - 27

d, ( 2x + y ) ( 4x² - 2xy + y² ) - 2xy ( 4x² + 2xy + y² )

= ( 2x )³ + y³ - 8x³y - 4x²y² - 2xy³ 

= 8x³ + y³ - 8x³y - 4x²y² - 2xy³

\(b,n^4-n^2=n^2\left(n^2-1\right)=n^2\left(n-1\right)\left(n+1\right)\)

\(=n.n\left(n-1\right)\left(n+1\right)\)

xét \(n=2k\)

\(n.n=4k⋮4\)

xét \(n=2k+1\)

\(\left(n-1\right)\left(n+1\right)=2k\left(2k+2\right)=4k\left(k+1\right)⋮4\)

\(< =>n.n\left(n-1\right)\left(n+1\right)⋮4\)

\(n^4-n^2⋮4< =>ĐPCM\)