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:

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

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

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

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

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

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

Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html

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

Lời giải:
$x^2+x-6=(x^2-2x)+(3x-6)=x(x-2)+3(x-2)=(x-2)(x+3)$

\(4x+8+x+2\\ =4\left(x+2\right)+1.\left(x+2\right)\\ =\left(x+2\right)\left(4+1\right)\)

Đề là +8 mới phân tích được nhé!

\(\text{4x + 8x + x + 2}\)

\(\text{= (4x + 8x + x) + 2}\)

\(\text{= 13x + 2}\)

Đa thức không phân tích được bạn ơi

Đề sai nhé .Sửu lại

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

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

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

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

\(\left(a+b\right).\left(b+c\right).\left(c-a\right)+\left(b+c\right).\left(c+a\right).\left(a-b\right)+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

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

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

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

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

giai lai:

\(\left(b+c\right).\left[\left(a+b\right).\left(c-a\right)+\left(c+a\right).\left(a-b\right)\right]+\left(c+a\right).\left(a+b\right).\left(b-c\right)\)

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

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

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

\(=\left(b-c\right).\left(a+b\right).\left(a-c\right)\)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

bạn ơi bạn chưa bớt 2x^2 kìa

ngu dốt

x⁵-x⁴-1=x⁵-x³-x²-x⁴+x²+x+x³-x-1

=x².(x³-x-1)-x.(x³-x-1)+(x³-x-1)

=(x³-x-1)(x²-x+1)

x^4+x^2+1 = (x^4+2x^2+1)-x^2 = (x^2+1)^2-x^2 = (x^2-x+1).(x^2+x+1)

k mk nha

mik chịu

\(x^8+x^4+1\)

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

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

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

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