(x-1)(x-2)(x+4)(x+5)-72=[(x-1)(x+4)][x-2)(x+5)]-72=(x^2+3x-4)(x^2+3x-10)-72

Đặt x^2+3x-4=t nên x^2+3x-10=t-6. Thay vào (*) ta được :

(x-1)(x-2)(x+4)(x+5)=t.(t-6)-72=t^2-6t-72=t^2-6t+9-81=(t-3)^2-9^2=(t-3-9)(t-3+9)=(t-12)(t+6)=(x^2+3x-16)(x^2+3x+2)

M = x⁹ - x⁷ + x⁶ - x⁵ - x⁴ + x³ - x² + 1

= ( x⁹ - x⁷ ) + ( x⁶ - x⁴ ) - ( x⁵ - x³ ) - ( x² - 1 )

= x⁷( x² - 1 ) + x⁴( x² - 1 ) - x³( x² - 1 ) - ( x² - 1 )

= ( x² - 1 )( x⁷ + x⁴ - x³ - 1 )

= ( x - 1 )( x + 1 )[ x⁴( x³ + 1 ) - ( x³ + 1 ) ]

= ( x - 1 )( x + 1 )( x³ + 1 )( x⁴ - 1 )

= ( x - 1 )( x + 1 )( x + 1 )( x² - x + 1 )( x² - 1 )( x² + 1 )

= ( x + 1 )²( x - 1 )( x² - x + 1 )( x - 1 )( x + 1 )( x² + 1 )

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

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\)

\(=4\left(x+5\right)\left(x+10\right)\left(x+12\right)\left(x+6\right)-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

\(=4\left(x+60\right)^2+132x\left(x+60\right)+1088x^2-3x^2\)

\(=4\left(x+60\right)^2+132x\left(x+60\right)+1085x^2\)

\(=4\left(x+60\right)^2+62x\left(x+60\right)+70x\left(x+60\right)+1085x^2\)

\(=2\left(x+60\right)\left[2\left(x+60\right)+31x\right]+35x\left[2\left(x+60\right)+31x\right]\)

\(=\left(33x+120\right)\left(2x+120+35x\right)\)

\(=3\left(11x+40\right)\left(37x+120\right)\)

Câu 1:

\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)-112\)

\(=\left(x-1\right)\left(x+4\right)\left(x-2\right)\left(x+5\right)-112\)

\(=\left(x^2+3x-4\right)\left(x^2+3x-10\right)-112\)

\(=\left(x^2+3x-7\right)^2-3^2-112\)

\(=\left(x^2+3x-7\right)^2-11^2\)

\(=\left(x^2+3x+4\right)\left(x^2+3x-18\right)\)

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

Câu 2:

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)-72\)

\(=\left(x^2-4\right)\left(x^2-10\right)-2\)

\(=\left(x^2-7\right)^2-3^2-72\)

\(=\left(x^2-7\right)^2-81\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

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

(x−1)(x−2)(x+4)(x+5)−112

=(x−1)(x+4)(x−2)(x+5)−112

=(x^2+3x−4)(x^2+3x−10)−112

=(x^2+3x−7)^2−32−112

=(x^2+3x−7)^2−112

=(x^2+3x+4)(x^2+3x−18)

=(x^2+3x+4)(x+6)(x−3)

Câu 2:

(x−2)(x+2)(x^2−10)−72

=(x2−4)(x^2−10)−2

=(x^2−7)^2−32−72

500,600,700,800,900,1000

cách làm :

500 thì bỏ 2 số 0 thì sẻ =5 tương tự 600=6 cứ nối tiếp 7,8,9

rồi thì trả lại số 0 cho các số 

nếu số đầu là 2 số 0 thì các số khác cũng sẽ 2 số 0

 vậy các số cần tìm là : 600, 700,800,900

chúc em học tốt dù hơi rối xíu

\(a,=4x^3\left(x+1\right)-x\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\\ =x\left(2x-1\right)\left(2x+1\right)\left(x+1\right)\\ b,=\left(a-1\right)^2-\left(b-c\right)^2\\ =\left(a-1-b+c\right)\left(a-1+b-c\right)\\ c,=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\\ =\left(x^2-9x+17\right)^2-9-72\\ =\left(x^2-9x+17\right)^2-81=\left(x^2-9x+8\right)\left(x^2-9x+26\right)\\ =\left(x-1\right)\left(x-8\right)\left(x^2-9x+26\right)\)

\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)+x^2\)

\(=4\left(x^2+5x+10x+50\right)\left(x^2+12x+6x+72\right)+x^2\)

\(=4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)+x^2\)

\(=4\left(x+6\right)\left(x+10\right)\left(x+5\right)\left(x+12\right)+x^2\)

\(=4\left(x^2+16x+60\right)\left(x^2+17x+60\right)+x^2\)

Đặt \(x^2+16x+60=a\)thay vào ta được :

\(4a\left(a+x\right)+x^2\)

\(=4a^2+4ax+x^2\)

\(=\left(2a+x\right)^2\)

\(=\left(2x^2+32x+120+x\right)^2\)

Chết lỡ enter rồi viết gọn nốt đoạn cuối giúp mình nhé :">