Bạn chuyển tất cả hạng tử từ vế phải sang vế trái ta được

\(^{x^2+5\text{x}^3+x^2y=5\text{x}^3+x^2y}\)

\(x^2+5\text{x}^3+x^2y-5\text{x}^3-x^2y=0\)

Rút gọn ta được

\(x^2=0\)

\(=>x=0\)

tick cho mình nha

x² - axy - bxy + aby² 

= ( x² - axy) - ( bxy - aby²)

= x( x-ay) - by( x - ay)

= ( x-ay)( x - by)

\(\left(x-5\right)\left(x-1\right)\left(x+3\right)\left(x+7\right)+60\)

\(=\left(x^2+2x-35\right)\left(x^2+2x-3\right)+60\)

\(=\left(x^2+2x\right)^2-38\left(x^2+2x\right)+105+60\)

\(=\left(x^2+2x\right)^2-3\left(x^2+2x\right)-35\left(x^2+2x\right)+165\)

\(=\left(x^2+2x-3\right)\left(x^2+2x-35\right)\)

\(=\left(x+3\right)\left(x-1\right)\left(x+7\right)\left(x-5\right)\)

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

\(=x^4+16x^2+9+8x^3-24x-6x^2-5x^2-20x+15+6x^2\)

\(=x^4+8x^3+11x^2-44x+24\)

\(=\left(x^4-x^3\right)+\left(9x^3-9x^2\right)+\left(20x^2-20x\right)-\left(24x-24\right)\)

\(=x^3\left(x-1\right)+9x^2\left(x-1\right)+20x\left(x-1\right)-24\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+9x^2+20x-24\right)\)

\(5x^2-4\left(x^2-2x+1\right)-5=\left(5x^2-5\right)-4\left(x-1\right)^2=5\left(x^2-1\right)-4\left(x-1\right)^2=5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=\left(x-1\right)\left(5x+5-4x+4\right)=\left(x-1\right)\left(x+9\right)\)

\(= \)\(5x^2-4x^2+8x-4-5\)

\(=\)\(x^2+8x-9\)

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

\(=(x-1)(x+9)\)

\(5x^2-4\left(x^2-2x+1\right)-5\)

\(=5x^2-4x^2+8x-4-5\)

\(=x^2+8x-9\)

\(=\left(x+9\right)\left(x-1\right)\)

\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15=\left(x^2+5x-3\right)\left(x^2+5x-3-2\right)-15=\left(x^2+5x-3\right)^2-2\left(x^2+5x-3\right)+1-16=\left(x^2+5x-3-1\right)^2-4^2=\left(x^2+5x-4\right)^2-4^2=\left(x^2+5x-8\right)\left(x^2+5x\right)=x\left(x+5\right)\left(x^2+5x-8\right)\)

\(\left(x^2+5x-3\right)\left(x^2+5x-5\right)-15\)

\(=\left(x^2+5x\right)^2-8\left(x^2+5x\right)-15\)

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

a) x²+x-2
= x²-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x²+5x+3
= 2x²+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x²+5x-2
= 3x²+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)

(x+3)(x+4)(x+5)(x+6)-24=[(x+3)(x+6)][(x+4)(x+5)]-24

                                  =(x²+6x+3x+3.6)(x²+5x+4x+5.4)-24

                                  =(x²+9x+18)(x²+9x+20)-24

                                  =(x²+9x+18)(x²+9x+18+2)-24    (*)

đặt x²+9x+18 là t   (1)

(*) trở thành

t(t+2)-24=t²+2t-24=t²-4t+6t-24

                          =(t²-4t)+(6t-24)

                          =t(t-4)+6(t-4)

                          =(t-4)(t+6)           (2)

thay (2) vào (1), ta được:

(x+3)(x+4)(x+5)(x+6)-24=(x²+9x+18-4)(x²+9x+18+6)

                                   =(x²+9x+14)(x²+9x+24)

                                   =(x²+7x+2x+14)(x²+9x+24)

                                   =[(x²+7x)+(2x+14)](x²+9x+24)

                                   =x(x+7)+2(x+7)(x²+9x+24)

                                   =(x+7)(x+2)(x²+9x+24)

(mình đã cố gắng giải thật chi tiết và phân tích triệt để nhất có thể rồi. có j sai sót thì góp ý nha!)

chả hiểu cái đề -_- 

( x + 3 )( x+ 4 )( x+  5 )( x+  6 ) - 24 

= ( x+ 3 )( x+ 6 )( x+ 4 )( x+ 5 ) - 24

 ( x^2 + 9x + 18 )( x^2 + 9x + 20 ) - 24 

Đặt x^2 + 9x + 19 = a 

= ( a - 1 )( a+ 1 ) - 24

= a^2 - 1 - 24 

= a^2 - 25 

= ( a- 5 )( a+ 5 ) 

= ( x^2 + 9x + 19 - 5 )( x^2 + 9x + 19 + 5 )

= ( x^2 + 9x + 14 )( x^2 + 9x + 24 ) 

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

x⁴ - x³ - x + 1

= (x⁴ - x³) - (x - 1)

= x³(x - 1) - (x - 1)

= (x³ - 1)(x - 1)