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\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2x^2.1+1^2\right]-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
\(\left(x^2-8\right)^2+36\)
\(=x^4-16x^2+64+36\)
\(=\left[\left(x^2\right)^2-2.10x^2+10^2\right]-\left(2x\right)^2\)
\(=\left(x^2-10\right)^2-\left(2x\right)^2\)
\(=\left(x^2-10-2x\right)\left(x^2-10+2x\right)\)
\(4x^4+81\)
\(=\left[\left(2x^2\right)^2+2.2x^2.9+9^2\right]-\left(6x\right)^2\)
\(=\left(2x^2+9\right)-\left(6x\right)^2\)
\(=\left(2x^2+9-6x\right).\left(2x^2+9+6x\right)\)
Tham khảo nhé~
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
Bài giải:
a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
Bài làm ai trên 11 điểm tích mình thì mình tích lại
Ông tùng hơn tùng số tuổi là :
29 + 32 = 61 (tuổi )
Vậy ông của tùng hơn tùng 61 tuổi
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
Mình xin lỗi nhé, để mình sửa lại : ^^
a) \(x^4+3x^2+4=\left(x^4+x^3+2x^2\right)+-\left(x^3+x^2+2x\right)+2\left(x^2+2x+2\right)\)
\(=x^2\left(x^2+x+2\right)-x\left(x^2+x+2\right)+2\left(x^2+x+2\right)=\left(x^2-x+2\right)\left(x^2+x+2\right)\)
b) \(x^4+5x^2+9=\left(x^4+x^3+3x^2\right)-\left(x^3+x^2+3x\right)+3\left(x^2+x+3\right)\)
\(=x^2\left(x^2+x+3\right)-x\left(x^2+x+3\right)+3\left(x^2+x+3\right)=\left(x^2-x+3\right)\left(x^2+x+3\right)\)
a) 9 -(x-y)2
= 32 - (x-y)2
= (3-x+y).(3+x-y)
b) (x2 +4)2 - 16x2
= (x2+4)2 - (4x)2
= (x2 + 4 -4x).(x2 + 4 +4x)
\(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
a) 4x2 - 36x + 56
= (2x)2 - 2 . 2x . 9 + 81 - 25
= ( 2x - 9 )2 - 52
= ( 2x - 9 + 5) ( 2x - 9 - 5 )
= ( 2x - 4 ) ( 2x - 14 )
= 4 ( x - 2 ) ( x - 7 )