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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)
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)
\(x^3-3x+2=x^3-x-2x+2=\left(x^3-x\right)-\left(2x-2\right)=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)=\left(x-1\right)\left[x\left(x+1\right)-2\right]\)
b)
\(\left(x^3+3x^2+3x+1\right)+\left(5x^2+10x+5\right)+\left(4x+4\right)\)
\(\left(x+1\right)^3+5\left(x+1\right)^2+4\left(x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2+5\left(x+1\right)+4\right]\)
4( x+5) ( x+6) (x+10) ( x+12) -3x2
=4(x+5)(x+12)(x+6)(x+10)-3x2
=4.(x2+17x+60)(x2+16x+60)-3x2
Đặt t=x2+16x+60 ta được:
4.(t+x).t-3x2
=4t2+4tx-3x2
=4t2-2tx+6tx-3x2
=2t.(2t-x)+3x.(2t-x)
=(2t-x)(2t+3x)
thay t=x2+16x+60 ta được:
[2.(x2+16x+ 60)-x][2.(x2+16x+60)+3x]
=(2x2+32x+120-x)(2x2+32x+120+3x)
=(2x2+31x+120)(2x2+35x+120)
=(2x2+16x+15x+120)(2x2+35x+120)
=[2x.(x+8)+15.(x+8)](2x2+35x+120)
=(x+8)(2x+15)(2x2+35x+120)
4( x+5) ( x+6) (x+10) ( x+12) -3x 2
=4(x+5)(x+12)(x+6)(x+10)-3x 2
=4.(x 2+17x+60)(x 2+16x+60)-3x 2
Đặt t=x 2+16x+60 ta được: 4.(t+x).t-3x 2
=4t 2+4tx-3x 2
=4t 2 -2tx+6tx-3x 2
=2t.(2t-x)+3x.(2t-x)
=(2t-x)(2t+3x)
thay t=x 2+16x+60 ta được: [2.(x 2+16x+ 60)-x][2.(x 2+16x+60)+3x]
=(2x 2+32x+120-x)(2x 2+32x+120+3x)
=(2x 2+31x+120)(2x 2+35x+120)
=(2x 2+16x+15x+120)(2x 2+35x+120)
=[2x.(x+8)+15.(x+8)](2x 2+35x+120)
=(x+8)(2x+15)(2x 2+35x+120)
a) x4 + 4 = (x4 + 4x2 + 4) - 4x2 = (x2 + 2)2 - 4x2 = (x2 + 2x + 2)(x2 - 2x + 2)
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24 = (x + 2)(x + 5)(x + 3)(x + 4) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = y => y(y + 2) - 24 = y2 + 2y - 24
= y2 + 6y - 4y - 24 = (y - 4)(y + 6) = (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16) = (x2 + x + 6x + 6)(x2 + 7x + 16) = (x + 1)(x + 6)(x2 + 7x + 16)
a) Đăt \(x^2+x=t\) khi đó bt trở thành:
\(t^2-2t-15=t^2+3t-5t-15=t\left(t+3\right)-5\left(t+3\right)\\ =\left(t+3\right)\left(1-5\right)=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
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 )
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left[\left(x^2+7x+11\right)-1\right]\left[\left(x^2+7x+11\right)+1\right]-24\)
\(=\left(x^2+7x+11\right)^2-1-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)