\(15x^2-34x+15\)

\(=15x^2-25x-9x+15\)

\(=5x\left(3x-5\right)-3\left(3x-5\right)\)

\(=\left(5x-3\right)\left(3x-5\right)\)

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

\(2ab^2-a^2b-b^3\)

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

\(=-b\left(a-b\right)^2\)

-(2ab² - a²b - b³)

= b(-2ab + a² + b²)

= b(a² - 2ab + b²)

= b(a - b)²

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

1.

= (x^3 + 125 ) -(x^2 +5x)

=(x +5) (x^2 -5x +25) -x(x+5)

=(x+5)(x^2 -5x +25 -x)

=(x+5)(x^2 -6x +25)

2.

= (x^3 -27) + (2x^2 -6x)

=(x-3) (x^2 +3x +9) +2x (x-3)

=(x-3) (x^2 +3x +9 +2x)

=(x-3) (x^2 +5x +9)

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)

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

\(=a^2b-ab^2+b^2c-bc^2+ac^2-a^2c\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a^2-bc-ab-ac\right)\)

\(=\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]\)

a)\(2x^2-12x=-18\)

\(\Leftrightarrow2x^2-12x+18=0\)

\(\Leftrightarrow2\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2\left(x-3\right)^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b) \(\left(4x^2-4x+1\right)-x^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(2x-1-x\right)\left(2x-1+x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)

_Minh ngụy_

\(x^2-ay-y^2-ax\)

\(=\left(x^2-y^2\right)-\left(ax+ay\right)\)

\(=\left(x-y\right)\left(x+y\right)-a\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-a\right)\)

_Minh ngụy_

\(a,\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)-\left(4x\right)^2=\left(x^2+4-4x\right).\left(x^2+4+4x\right)=\left(x-2\right)^2.\left(x+2\right)^2\)

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

\(c,\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right).\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right).\left(8x^2-18\right)\)