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Câu A hình như sai đề. nếu sai => sửa đề => ib = làm
b) \(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(B=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+10=y\)
\(\Rightarrow B=y.\left(y+2\right)-24\)
\(B=y^2+2y-24\)
\(B=\left(y^2+2y+1\right)-25\)
\(B=\left(y+1\right)^2-5^2\)
\(B=\left(y-4\right)\left(x+6\right)\)
\(B=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
Tham khảo nhé~
\(A=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-20\)
\(=\left(x^2+5x+4\right)\cdot\left(x^2+5x+6\right)-20\)
Đặt: \(x^2+5x+5=a\)Khi đó ta có:
\(A=\left(a-1\right)\left(a+1\right)-20=a^2-21=\left(a-\sqrt{21}\right)\left(a+\sqrt{21}\right)\)
tự thay trở lại
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) đề thế này\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)(1)
Đặt \(x^2+7x+11=t\)vào (1) ta được:
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)Thay \(t=x^2+7x+11\)ta được:
\(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
b) Phân tích sẵn rồi còn phân tích gì nưa=))
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)( Làm đề theo Lê Tài Bảo Châu )
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left[\left(x^2+7x+11\right)-1\right]\left[\left(x^2+7x+11\right)+1\right]-24\)
\(=\left(x^2+7x+11\right)^2-1-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
\(x^4-3x^3+5x^2-9x+6\)
\(=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\)
\(=x^3\left(x-2\right)-x^2\left(x-2\right)+3x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3-x^2+3x-3\right)\)
\(=\left(x-2\right)\left[x^2\left(x-1\right)+3\left(x-1\right)\right]\)
\(=\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\)
\(x^4+3x^2y^2+4y^4\)
\(x^4+4y^4-2xy^3+2xy^3+2x^2y^2+2x^2y^2-x^2y^2\)
\(+x^3y-x^3y\)
\(=\left(4y^4-2xy^3+2x^2y^2\right)+\left(2xy^3-x^2y^2+x^3y\right)\)
\(+\left(2x^2y^2-x^3y+x^4\right)\)
\(=2y^2\left(2y^2-xy+x^2\right)+xy\left(2y^2-xy+x^2\right)\)
\(+x^2\left(2y^2-xy+x^2\right)\)
\(=\left(2y^2+xy+x^2\right)\left(2y^2-xy+x^2\right)\)
a)18x2-12x
=3x(6x-4)
b)3x2-11x+6
=x(3x-11+6)
=x(3x-5)
c)x3+6x2+11x+6
=x2(x+23
\(18x^2-12x\)
\(=6x\left(3x-2\right)\)
\(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)