\(4x^3-13x^2+9x-18 \)

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

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

x²⁺y²-x²y²+xy-x-y=x²-x²y²+y²-y-x+xy

                            =x(1-y²)+y(y-1)-x(1-y)

                            =x²(y-1)(y+1)+y(y-1)+x(y-1)

                           =-x²(y-1)(y+1)+y(y-1)+x(y-1)

                           =(y-1)(-x²(y+1)+y+x)

f)    x⁴+2x²-4x-4=(x³.x+x³.2)-(2x.2+2.2)

                          =x³(x+2)-2(x+2)

                            =(x³-2)(x+2)

c, \(2x^2+x-3=x\left(2x-3\right)\)

d, \(6x^2-x-15=x\left(6x-15\right)\)

TK MIK NHA

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

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

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

Tham khảo nhé~

a. \(x^2-x+\frac{1}{4}=x^2-2x\frac{1}{2}+\left(\frac{1}{2}\right)^2=\left(x-\frac{1}{2}\right)^2\)

b. \(4x^2-4x+1=\left(2x\right)^2-2.2x.1+1^2=\left(2x-1\right)^2\)

c. \(x^2-3x+\frac{9}{4}=x^2-2x\frac{3}{2}+\left(\frac{3}{2}\right)^2=\left(x-\frac{3}{2}\right)^2\)

TK MIK NHA~

\(C=x^2-2xy+y^2+4y^2+4y+1+2=\left(x-y\right)^2+\left(2y+1\right)^2+2\ge2\)

Dấu "=" xảy ra khi\(\hept{\begin{cases}x-y=0\\2y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y\\y=\frac{-1}{2}\end{cases}\Leftrightarrow}x=y=\frac{-1}{2}}\)

Câu hỏi của TH - Toán lớp 8 - Học toán với OnlineMath tham kahr