Lời giải:
$(2x+3)^2+(2x+5)^2=4x^2+12x+9+4x^2+20x+25$

$=8x^2+32x+34$

\(x^3+3x+7x^3+2x^2+5x+1\)

\(=8x^3+2x^2+8x+1\)

tham khảo nhé~

=(x^3+7x^3)+2x^2+(3x+5x)+1

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

Hok tot!!!!

7x³ + x³ + 2x² + 5x + 3x + 1

= (7x³ + x³) + 2x² + (5x + 3x) + 1

= 8x³ + 2x² + 8x + 1

= 2x.(4x² + x + 4) + 1

Chúc em học tốt!!!

N=2(2x + 5 )^2 - 3(1 + 4x )(1 - 4x)

= 2 (4x^2 + 20x + 25) - 3(1 - 16x^2)

= 8x^2 + 40x + 50 - 3 + 48x^2

= 56x^2 + 40x - 47

(.....????!!!!!!.....)

help voi a

a: \(\left(3x+2\right)^2+4x-3x^2+2\left(5x-2\right)\left(5x+2\right)-75x^2\)

\(=9x^2+12x+4+4x-3x^2+50x^2-8-75x^2\)

\(=-19x^2+16x-4\)

Ta có 2x(2x + 1)² - 3x(x + 3)(x - 3) - 4x(x + 1)²

= 2x(4x² + 4x + 1) - 3x(x² - 9) - 4x(x² + 2x + 1) 

= 8x³ + 8x² + 2x - 3x³ + 27x - 4x³ - 8x² - 4x

= 8x³ - 3x³ - 4x³ + 8x² - 8x² + 2x + 27x - 4x

= x³ + 25x 

a oi hinh nhu sai r con +16x² nua co a , anh tinh lai ho e duoc kh

Kiều Trinh Vũ ko có + 16x² nhá vì 8x² - 8x² = 0 nhá 

\(:P=5x(x-3)(x+3)-(2x-3)^2+34x(x+2)-5(x+2)^3+25x-1\)

\(P=5x(x^2-9)-(4x^2-12x+9)+34x^2+68x-5(x^3+6x^2+12x+8)+25-1\)

\(P=5x^3-45x-4x^2+12x-9+34x^2+68x-5x^3-30x^2-60x-40+25-1\)

\(P=(5x^3-5x^3)+(34x^2-4x^2-30x^2)+(12x-45x++68x+25x-60x)-(9+1)\)

\(P=-10\)