\(=\left(3x+5-3x+5\right)^2=100\)

Bài 1 :

\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)

\(B=27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)

Vì \(100< 104\Rightarrow A< B\)

Bài 2 :

\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

\(\Rightarrow4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)

\(\Rightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)

\(\Rightarrow4x=-2\)\(\Leftrightarrow x=-\frac{1}{2}\)

Chia nhỏ ra ik ạ

\(\left(2x+3\right)\left(2x-3\right)-4x\left(x+5\right)=4x^2-9-4x^2-20x=-20x-9\)

\(5x\left(x-3\right)+\left(x-2\right)^2=5x^2-15x+x^2-4x+4=6x^2-19x+4\)

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

a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2 

= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25

= 36

b) (3x^2 - y)^2

= 9x^4 - 6x^2y + y^2

c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)

= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4

= 9x^2 + 54

d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2

= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x

= x^3 - 16x^2 + 25x

e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)

= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2

= x^3 + 2x^2 - 2x - 12

f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2

= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4

= x^6 + 2x^4 + 2x^2 + 124

P(x)+Q(x)

=3x^2y-2x+5xy^2-7y^2+3xy^2-7y^2-9x^2y-x-5

=8xy^2-14y^2-6x^2y-3x-5

=>Chọn A

a: =3x^3-15x^2+21x

b: =-x^3+6x^2+5x-4x^2-24x-20

=-x^3+2x^2-19x-20

c: =9x^2+15x-3x-5-7x^2-14

=2x^2+12x-19

d: =10x^2-4x+2/3

\(Q-P=3x^3-3x^2+8x-5-5x^2+3x-2\)

\(Q-P=3x^3+\left(-3x^2-5x^2\right)+\left(8x+3x\right)+\left(-5-2\right)\)

\(Q-P=3x^3-8x^2+11x-7\)

Bài 1.

[ 4( x - y )⁵ + 2( x - y )³ - 3( x - y )² ] : ( y - x )² < sửa một lũy thừa rồi nhé >

= [ 4( x - y )⁵ + 2( x - y )³ - 3( x - y )³ ] : ( x - y )²

Đặt t = x - y

bthuc ⇔ ( 4t⁵ + 2t³ - 3t² ) : t²

           = 4t⁵ : t² + 2t³ : t² - 3t² : t²

           = 4t³ + 2t - 3

           = 4( x - y )³ + 2( x - y ) - 3

Bài 2.

5x( x - 2 ) + 3x - 6 = 0

⇔ 5x( x - 2 ) + 3( x - 2 ) = 0

⇔ ( x - 2 )( 5x + 3 ) = 0

⇔ x - 2 = 0 hoặc 5x + 3 = 0

⇔ x = 2 hoăc x = -3/5

Bài 3.

A = x² - 6x + 2023

= ( x² - 6x + 9 ) + 2014

= ( x - 3 )² + 2014 ≥ 2014 ∀ x

Dấu "=" xảy ra khi x = 3

=> MinA = 2014 <=> x = 3

Bài 4.

B = ( 3x + 5 )² + ( 3x - 5 )² - 2( 3x + 5 )( 3x - 5 )

= [ ( 3x + 5 ) - ( 3x - 5 ) ]²

= ( 3x + 5 - 3x + 5 )²

= 10² = 100

Vậy B không phụ thuộc vào x ( đpcm )

Bài 6.

C = 1² - 2² + 3² - 4² + 5² - 6² + ... + 2013² - 2014² + 2015²

= ( 2015² - 2014² ) + ... + ( 5² - 4² ) + ( 3² - 2² ) + 1

= ( 2015 - 2014 )( 2015 + 2014 ) + ... + ( 5 - 4 )( 5 + 4 ) + ( 3 - 2 )( 3 + 2 ) + 1

= 4029 + ... + 9 + 5 + 1

= \(\frac{\left(4029+1\right)\left[\left(4029-1\right)\div4+1\right]}{2}\)

= 2 031 120