(x+2)(x+3)(x+4)(x+5)-24

= [(x+2)(x+5)][(x+3)(x+4)] -24

=(x^2+7x+10)(x^2+7x+12)-24

thay x^2+7x+11=y

=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)

= (x^2+7x+11-5)(x^2+7x+11+5)=(x^2+7x+6)(x^2+7x+16)=(x^2+x+6x+6)(x^2+7x+16)=[x(x+1)+6(x+1)]((x^2+7x+16)=(x+1)(x+6)(x^2+7x+16)

(x + 2)(x + 3)(x + 5)(x + 7) - 24

= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24

=(x² + 7x + 10)(x² + 7x +12) - 24

Đặt x² + 7x + 11 = t ; ta có:

(t - 1)(t + 1) - 24

= t² - 1² - 24

= t² - 25

= (t - 5)(t + 5)

Thay t = x² + 7x + 11 ta được:

(x² + 7x + 11 - 5)(x² + 7x +11 + 5)

= (x² + 7x + 6)(x² + 7x + 16)

= (x + 1)(x + 6)(x² + 7x + 16)

Chúc bn học tốt

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)( * )

Đặt \(t=x^2+7x+10\), khi đó (*) trở thành:

\(t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

Thay \(t=x^2+7x+10\) vào, ta được:

\(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

Vậy ...

a) x³ + 4x² - 29x + 24                                                           

= x³ - 3x² + 7x² - 21x - 8x + 24

= x²(x-3) + 7x(x-3) - 8(x-3)

= (x-3)(x²+7x-8)

=(x-3)(x²+8x-x-8)

= (x-3)[(x²+8x)-(x+8)]

= (x-3)[x(x+8)-(x+8)]

= (x-3)(x+8)(x-1)

a) =x³-2x²+6x²-12x -12x +24

= x²(x-2)+6x(x-2)-12(x-2)

= (x-2)(x²+6x-12)

mk giải đc câu a thôi, bn zô jup mk lại vs

\(a,x^3+4x^2-24x+24\)

\(=x^3+6x^2-12x-2x^2-12x+24\)

\(=\left(x^3-2x^2\right)+\left(6x^2-12x\right)-\left(12x-24\right)\)

\(=x^2\left(x-2\right)+6x\left(x-2\right)-12\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+6x-12\right)\)

b)

x⁸+x⁴+1

= x⁸+x⁷+x⁶-x⁷-x⁶-x⁵+x⁵+x⁴+x³-x³-x²-x+x²+x+1

= x⁶(x²+x+1)-x⁵(x²+x+1) +x³(x²+x+1)-x(x²+x+1)+(x²+x+1)

= (x²+x+1)(x⁶-x⁵+x³-x+1)

= (x²+x+1)(x⁶-x⁵+x⁴-x⁴+x³-x²+x²-x+1)

= (x²+x+1)[x⁴(x²-x+1) - x²(x²-x+1) + (x²-x+1)]

= (x²+x+1)(x²-x+1)(x⁴-x²+1)

\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

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

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

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

Giải giùm em \(\left(x^2+4x+8\right)^2+3x^3+14x^2+24x\) nha

\(=\left(a-1\right)\left(a+4\right)\left(a+3\right)\left(a-2\right)-24=\left(a-2\right)\left(a+4\right)\left(a-1\right)\left(a+3\right)-24\)\(=\left(a^2+2a-8\right)\left(a^2+2a-3\right)-24.dat:a^2+2a-8=h\)\(\Rightarrow\left(a^2+2a-8\right)\left(a^2+2a-3\right)-24=h\left(h+5\right)-24=h^2+5h-24=\left(h-3\right)\left(h+8\right)\)\(=\left(a^2+2a-11\right)a\left(a+2\right)\)

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

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

\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)

\(=\left(x^2+7x+11\right)^2-9\)

\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

Bài 1:

a) \(x^2-2xy-25+y^2\) (Sửa đề)

\(=x^2-2xy+y^2-25\)

\(=\left(x-y\right)^2-5^2\)

\(=\left(x-y-5\right)\left(x-y+5\right)\)

Vậy ...

b) \(x\left(x-1\right)+y\left(1-x\right)\)

\(=x\left(x-1\right)-y\left(x-1\right)\)

\(=\left(x-1\right)\left(x-y\right)\)

Vậy ...

c) \(7x+7y-\left(x+y\right)\) (Sửa đề)

\(=7\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(7-1\right)\)

\(=6\left(x+y\right)\)

Vậy ...

d) \(x^4+y^4\)

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

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

\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)

Vậy ...

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