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

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)

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

a: \(2y\left(x+2\right)-3x-6\)

\(=2y\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2y-3\right)\)

b: \(3\left(x+4\right)-x^2-4x\)

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

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(x+4\right)\left(3-x\right)\)

c: \(2\left(x+5\right)-x^2-4x\)

\(=2x+10-x^2-4x\)

\(=-x^2-2x+10\)

\(=-x^2-2x-1+11\)

\(=11-\left(x^2+2x+1\right)\)

\(=11-\left(x+1\right)^2\)

\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)

d: \(x^2+6x-3x-18\)

\(=\left(x^2+6x\right)-\left(3x+18\right)\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x+6\right)\left(x-3\right)\)

x⁴ - 4x³ - 8x² + 8x 

 = x(x³ - 4x² - 8x + 8) 

= x[x³ + 8 - 4x(x + 2)] 

= x[(x + 2)(x² - 2x + 4) - 4x(x + 2)] 

= x(x + 2)(x² - 6x + 4)

= x(x + 2)(x² - 6x + 9 - 5) 

 = \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)

\(x^4-4x^3-8x^2+8x\)

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

\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)

\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)

\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)

\(=x\left(x^2-6x+4\right)\left(x+2\right)\)

\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)

\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)

x²(1-x²)-4-4x²

=x²(1-x²)-4(1-x²)

=(x²-4)(1-x²)

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

nhầm

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

\(=\left(x+2\right)^2\) bạn nào mới học mà không biết làm bài này thì chịu thua

x⁴+2x³+5x²+4x-12

=x(x³+2x²+5x+4)-12

Có thể chi tiết ra ko bạn, mình cảm ơn.

\(x^4+2x^3+5x^2+4x-12\)

\(=\left(x^4+2x^3+x^2\right)+4x^2+4x-12\)

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)

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

\(=\left(x^2+x\right)\left(x^2+x-2\right)+2\left(x^2+x-2\right)\)

\(=\left(x^2+x+2\right)\left(x^2+x-2\right)\)

\(=\left(x^2+x+2\right)\left(x^2-x+2x-2\right)\)

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(x^3+4x^2+4x-16y^2\)

\(=\left(x^3+2x^2\right)+\left(2x^2+4x\right)-16y^2\)

\(=x^2.\left(x+2\right)+2x.\left(x+2\right)-16y^2\)

\(=\left(x+2\right).\left(x^2+2x\right)-16y^2\)

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

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

\(=\left[\sqrt{x}.\left(x+2\right)\right]^2-4y^2\)

\(=\left[\sqrt{x}.\left(x+2\right)-4y\right].\left[\sqrt{x}.\left(x+2\right)+4y\right]\)

Tham khảo nhé~

nếu đưa vô căn phải có điều kiện là x > 0

\(x^3+4x^2+4x-16y^2=x\left(x+2\right)^2-\left(4y\right)^2\)

\(=\left(x\sqrt{x}+2\sqrt{x}\right)^2-\left(4y\right)^2=\left(x\sqrt{x}+2\sqrt{x}-4y\right)\left(x\sqrt{x}+2\sqrt{x}+4y\right)\)