A/\(\left(2x^3+y^2-7xy\right)4xy^2.\)

\(=8x^4y^2+4xy^4-28x^2y^3\)

B/\(\left(2x^3-x-1\right)\left(5x-2\right)\)

\(=10x^4-5x^2-5x-4x^3+2x+2\)

\(=10x^4-5x^3-3x-4x^3+2\)

C/\(\left(2x^2-3\right)\left(4x^4+6x^2+9\right)\)

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

D/\(\left(3x^2-2y\right)^3-\left(2x^2-y\right)^3\)

( Bài này áp dụng hằng đẳng thức là làm được ạ )

Đặt: 

\(x^3+ax+b=\left(x+1\right)q\left(x\right)+7\left(1\right)\)

\(x^3+ax+b=\left(x-3\right)p\left(x\right)-5\left(2\right)\)

Thay x = -1 và x = 3 lần lượt vào (1) và (2), ta có: 

\(\hept{\begin{cases}-1-a+b=7\\27+3a+b=-5\end{cases}\Rightarrow\hept{\begin{cases}-a+b=8\\3a+b=-32\end{cases}\Rightarrow}\hept{\begin{cases}a=-10\\b=-2\end{cases}}}\)

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

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

Mà \(\left(x-2\right)^2\ge0\) với mọi x \(\Rightarrow\left(x-2\right)^2-9\ge-9\) với mọi x

Vậy GTNN của  \(\left(x+1\right)\left(x-5\right)=-9\)đạt được khi \(x=2\)

1. Thực hiện phép chia đa thức: ta có kết quả:

\(x^3+5x^2+3x+a=\left(x+3\right)\left(x^2+2x+b\right)+\left(-3-b\right)x+a-3b\)

Để f(x) chia hết cho x²+2x+b thì -3-b=0 và a-3b=0 <=> b=-3; a=-9

      \(3a^2+4b^2=7ab\)

\(\Rightarrow3a^2+4b^2-7ab=0\)

\(\Rightarrow3a^2-3ab-4ab+4b^2=0\)

\(\Rightarrow3a\left(a-b\right)-4b\left(a-b\right)=0\)

\(\Rightarrow\left(a-b\right)\left(3a-4b\right)=0\)

Mà \(a\ne b\Rightarrow a-b\ne0\) 

Từ đó \(3a-4b=0\Rightarrow3a=4b\Rightarrow a=\frac{4}{3}b\)

\(E=\frac{a+2b}{3a-b}=\frac{\frac{4}{3}b+2b}{3.\frac{4}{3}b-b}=\frac{10}{9}\)

\(A=4x^2+10y^2-4xy-32y+4x+27\)       

\(=\left(4x^2-4xy+y^2\right)+4x-2y+1+9y^2-30y+25+1\)

\(=\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(3y\right)^2-2.3y.5+5^2+1\)

\(=\left(2x-y+1\right)^2+\left(3y-5\right)^2+1>0\forall x;y\)

Pham Van Hung

A=4x^2+10y^2-4xy-32y+4x+27A=4x2+10y2−4xy−32y+4x+27       

=\left(4x^2-4xy+y^2\right)+4x-2y+1+9y^2-30y+25+1=(4x2−4xy+y2)+4x−2y+1+9y2−30y+25+1

=\left(2x-y\right)^2+2\left(2x-y\right)+1+\left(3y\right)^2-2.3y.5+5^2+1=(2x−y)2+2(2x−y)+1+(3y)2−2.3y.5+52+1

=\left(2x-y+1\right)^2+\left(3y-5\right)^2+1&gt;0\forall x;y=(2x−y+1)2+(3y−5)2+1>0∀x;y

     \(a^3+b^3=3ab-1\)

\(\Rightarrow a^3+b^3+1-3ab=0\)

\(\Rightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b\right)=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)

Mà \(a,b>0\Rightarrow a+b+1>0\)

\(\Rightarrow a^2-ab+b^2-a-b+1=0\)

\(\Rightarrow2a^2-2ab+2b^2-2a-2b+2=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)

\(\Rightarrow a=b=1\Rightarrow a^{2018}+b^{2019}=1+1=2\)

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

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

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

                                      \(=\frac{x^2+6x-9}{x\left(x-3\right)\left(x+3\right)}\)

\(\frac{x+9}{x^2-9}-\frac{3}{x^2-3x}\)

\(=\frac{x+9}{\left(x+3\right)\left(x-3\right)}-\frac{3}{x\left(x-3\right)}\)

\(=\frac{\left(x+9\right)x}{x\left(x+3\right)\left(x-3\right)}-\frac{3\left(x+3\right)}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{x^2+9x-3x-9}{x\left(x+3\right)\left(x-3\right)}=\frac{x^2+6x-9}{x\left(x+3\right)\left(x-3\right)}\)

\(=\frac{\left(x+3\right)^2}{x\left(x+3\right)\left(x-3\right)}=\frac{x+3}{x\left(x-3\right)}\)

hok tốt ...