\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

a) xy+3x-7y-21

=x(y+3)-7(x+3)

=(x-7)(y+3)

b)2xy-15-6x-5y

=2x(y-3)-5(-3+y)

=(2x-5)(y-3)

c)2x^2y+2xy^2-2x-2y

=2x(xy-1)+2y(xy-1)

=(2x+2y)(xy-1)

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

\(a,=x^2-9-x^2+6x-9=6x-18\\ b,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x+y-5\right)\left(x-y\right)\)

\(x^3-9x^2+x=x\left(x^2-9x+1\right)\)

\(x^3+13x^2+x=x\left(x^2+13x+1\right)\)

\(b,x^3-2x^2-4xy^2+x\)

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

\(=x\left[\left(x^2-2x+1\right)-4y^2\right]\)

\(=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]\)

\(=x\left(x-1-2y\right)\left(x-1+2y\right)\)

\(=x\left(x-2y-1\right)\left(x+2y-1\right)\)

\(---\)

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

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

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

Đặt \(y=x^2+7x+10\), thay vào (1) ta được:

\(y\left(y+2\right)-8\)

\(=y^2+2y+1-9\)

\(=\left(y+1\right)^2-3^2\)

\(=\left(y+1-3\right)\left(y+1+3\right)\)

\(=\left(y-2\right)\left(y+4\right)\)

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

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

#Ayumu

x⁴ + x³ + 2x² + x + 1 

= (x⁴ + 2x² + 1) + (x³ + x)

= (x² + 1)² + x (x² + 1)

= (x² + 1) ( x² + 1 + x)

= (x² + 1) (x + 1)²

Trả lời:

x⁴ + x³ + 2x² + x + 1

= ( x⁴ + 2x² + 1 ) + ( x³ + x )

= ( x² + 1 )² + x ( x² + 1)

= ( x² + 1 ) ( x² + 1 + x )

= ( x² + 1) ( x + 1 )²

Bài 1:

\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)

Bài 3:

\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)

Bài 1;

1) \(x^3-2x-x=x\left(x^2-2x-1\right)\)

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

3) \(2y^3+8y^3+8y=10y^3+8y=2y\left(5y^2+4\right)\)

4) \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

Bài 2:

1) \(x^3-64x=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\)

2) \(8x^2y-18y=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\)

3) \(24x^3-3=3\left(8x^3-1\right)=3\left(2x-1\right)\left(4x^2+2x+1\right)\)

Bài 3:

1) \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y\right]=5\left(x-y+1\right)\left(x+y+1\right)\)

2) \(3x^3-6x^2+3x-12xy^2=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=3x\left(x-2y-1\right)\left(x+2y-1\right)\)

3) \(a^3b-ab^3+a^2+2ab+b^2=ab\left(a^2-b^2\right)+\left(a+b\right)^2=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(a^2b-ab^2+a+b\right)\)

4) \(2x^3-2xy^2-8x^2+8xy=2x\left(x^2-y^2-4x+4y\right)=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]=2x\left(x-y\right)\left(x+y-4\right)\)

a) Ta thấy đa thức \(f\left(x\right)=4x^2+81\) vô nghiệm (*).

 Giả sử \(f\left(x\right)\) có thể phân tích được thành nhân tử, khi đó \(f\left(x\right)=\left(ax+b\right)\left(cx+d\right)\), suy ra \(f\) có nghiệm là \(x=-\dfrac{b}{a}\) hoặc \(x=-\dfrac{d}{c}\), mâu thuẫn với (*).

 Vậy ta không thể phân tích \(f\left(x\right)\) thành nhân tử.

b) \(g\left(x\right)=x^7+x^2+1\)

\(g\left(x\right)=x^7-x+x^2+x+1\)

\(g\left(x\right)=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(g\left(x\right)=x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

\(g\left(x\right)=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

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

 Xét \(h\left(x\right)=x^5-x^4+x^2-x+1\), nếu \(h\left(x\right)\) phân tích được thành nhân tử thì nó có nghiệm hữu tỉ. Khi đó nó có dạng \(x=\dfrac{p}{q},\left(p,q\inℤ;\left(p,q\right)=1\right),p|1,q|1\) \(\Rightarrow x=\pm1\). Ta thấy \(h\left(1\right).h\left(-1\right)\ne0\) nên 2 nghiệm này không thỏa mãn. Vậy h(x) không có nghiệm hữu tỉ \(\Rightarrow\) g(x) không thể phân tích tiếp.

a)

\(4x^2+81\\=(2x)^2+2\cdot2x\cdot9+9^2-36x\\=(2x+9)^2-36x\)

Bạn xem lại đề bài nhé!

b)

\(x^7+x^2+1\\=(x^7+x^6+x^5)-x^6-x^5-x^4+(x^4+x^3+x^2)-(x^3-1)\\=x^5(x^2+x+1)-x^4(x^2+x+1)+x^2(x^2+x+1)-(x-1)(x^2+x+1)\\=(x^2+x+1)(x^4-x^4+x^2-x+1)\)