Lời giải:

$(x^3-2xy+3x^2)(x-2y)=x(x^2-2xy+3x)(x-2y)$

−5𝑥^2−8𝑥+4

\(=\left[2\left(x-1\right)-3x\right]\left[2\left(x-1\right)+3x\right]\\ =\left(2x-2-3x\right)\left(2x-2+3x\right)\\ =\left(-2-x\right)\left(5x-2\right)\\ =\left(x+2\right)\left(2-5x\right)\)

\(4\left(x-1\right)^2-9x^2=\left(2x-2\right)^2-\left(3x\right)^2=\left(2x-2-3x\right)\left(2x-2+3x\right)=\left(-x-2\right)\left(5x-2\right)=-\left(x+2\right)\left(5x-2\right)=\left(x+2\right)\left(2-5x\right)\)

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

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)

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

Câu 1:

\(4x^2+16x-9\)

\(=4x^2+18x-2x-9\)

\(=2x\left(2x+9\right)-\left(2x+9\right)\)

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

Câu 2:

\(6x^2-11x+3=0\)

\(\Leftrightarrow6x^2-2x-9x+3=0\)

\(\Leftrightarrow2x\left(3x-1\right)-3\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

a: =(x^2-1)^2-2x(x^2-1)+x(x^2-1)-2x^2

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

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

b: \(=\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2\)

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

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

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

\(\text{a) }4x^{16}+81=4x^4+36x^2+81-36x^8\)

                          \(=\left(4x^{16}+36x^8+81\right)-36x^8\)

                          \(=\left[\left(2x^8\right)^2+2.2x^8.9+9^2\right]+\left(6x^4\right)^2\)

                          \(=\left(2x^8+9\right)^2-\left(6x^4\right)^2\)

                         \(=\left(2x^8+9-6x^4\right)\left(2x^8+9+6x^4\right)\)                    

\(\text{b) }x^4+2018x^2+2017x+2018\)

\(=x^4+2018x^2+2018x-x+2018\)

\(=\left(x^4-x\right)+\left(2018x^2+2018x+2018\right)\)

\(=x\left(x^3-1\right)-2018\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)

\(=\left(x^2-x\right)\left(x^2+x+1\right)+2018\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2018\right)\)

Câu 1:

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

\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)