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Câu 1 :
a) \(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2+2x\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)
b) \(a^4+a^2+1\)
\(=\left(a^2\right)^2+2a^2+1-a^2\)
\(=\left(a^2+1\right)-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
c) \(x^4+64\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Câu 2 :
a) \(\left(a-b\right)^2=a^2-2ab+b^2\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)
\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)
b) tương tự
\(a^4+8a^3+14a^2-8a-15\)
\(=a^4+8a^3+15a^2-a^2-8a-15\)
\(=a^2\left(a^2+8a+15\right)-\left(a^2+8a+15\right)\)
\(=\left(a^2+8a+15\right)\left(a^2-1\right)\)
\(=\left(a+3\right)\left(a+5\right)\left(a-1\right)\left(a+1\right)\)
a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)
=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)
b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
=\(a^2b^2-2ab+1+a^2+2ab+b^2\)
=\(a^2b^2+a^2+b^2+1\)
=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
=\(\left(b^2+1\right)\left(a^2+1\right)\)
c,\(x^4+2x^3+2x^2+2x+1\)
=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)
=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
=\(\left(x+1\right)^2\left(x^2+1\right)\)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
\(\left(x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)\)
\(\Leftrightarrow2.2x=4x\)
p/s tham khảo nha
\(a^2-b^2-a+b\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)\)
p/s tham khảo
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)
\(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3+3x^2+8x+12\right)=\left(x-1\right).\left(x+2\right).\left(x^2+x+6\right)\)
p/s: sai sót bỏ qua