Ta có: 72⁷⁵ = 72^{4 x 18 + 3} = ( 72 ⁴ )¹⁸ x 72³ = (....6)¹⁸ x 72³ = (...6) x (...8) = (...8)

128¹²⁹ = 128^{4 x 32 + 1} = (128⁴)³² x 128 = (...6)³² x 128 = (...6) x 128 = (...8)

(234⁵)⁴² = 234^{5 x 42} = 234²¹⁰ = 234^{4 x 52 + 2} = (234⁴)⁵² x 234² = (...6)⁵² x 234² = (...6) x (...6) = (...6)

(579⁶)³⁵ = 579^{6 x 35} = 579²¹⁰= 579^{4 x 5 + 2} = (579⁴)⁵² x 579² = (...1)⁵² x (...8 ) = (...1) x (...8) = (...8)

HT~

`3x^2+4x=2x`

`<=>3x^2+2x=0`

`<=>x(3x+2)=0`

`<=>x=0,3x+2=0`

`<=>x=0,x=(-2)/3`

Vậy `x=0,x=(-2)/3`

help me

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

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

xin tiick

`(x^2-2xy+y^2)-9`

`=(x-y)^2-3^2`

`=(x-y-3)(x-y+3)`

\((3x-2).(3x+2)-4x.(2x+3)-(2x-1)^2\)

\(=(3x)^2-2^2-8x^2-12x-[(2x)^2-2.2x.1+1^2]\)

\(=9x^2-4-8x^2-12x-4x^2+4x-1\)

\(=(9x^2-8x^2-4x^2)+(4x-12x)-(4+1)\)

\(=-3x^2-8x-5\)

\((x+2).(x^2+3x+1)-(x+1).(x^2-x+1)+(x-2)^3\)

\(=x^3+3x^2+x+2x^2+6x+2-(x^3-1^3)+(x^3-3x^2.2+3x.2^2-2^3)\)

\(=x^3+3x^2+x+2x^2+6x+2-x^3-1+x^3-6x^2+12x-8\)

\(=(x^3-x^3+x^3)+(3x^2+2x^2-6x^2)+(x+6x+12x)+(2-1-8)\)

\(=x^3-x^2+19x-7\)

ta có: 

ta có :

Mình cần gấp lắm rồi

Từ a⁴ + b⁴ \(\ge\)2a²b² cộng a² + b² vào 2 vế

\(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2\)

Tương tự\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)

Từ đó suy ra \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^2\)

Cái cuối là \(a^4+b^4\ge\frac{1}{8}\left(a+b\right)^4\)nha mình nhầm