a,   2x^2 + 3x + 3 2x - 1 x + 2 2x^2 - x 4x + 3 4x - 2 5

b, Để giá trị đa thức A chia hết cho giá trị đa thức B <=> 

\(2x-1\inƯ\left(5\right)=\left\{1;5\right\}\)

a)\(\frac{x^2+xy}{x^2-y^2}=\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x}{x-y}\)

b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

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

a, x(x-y)+2(x-y)=(x-y)(x+2)

b, \(x^2-6xy+9y^2=\left(x-3y\right)^2\)Thay x=16, y=2 có

                \(x^2-6xy+9y^2=\left(x-3y\right)^2=\left(16-2\cdot3\right)^2=10^2=100\)

a³ + b³ + c³ = 3abc

⇔ ( a³ + b³ ) + c³ - 3abc = 0

⇔ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇔ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇔ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇔ ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

⇔ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇔ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Từ đây tự làm tiếp nhé :))

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

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

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

\(\Rightarrow\left(a+b+c\right)[\left(a+b+c\right)^2-3ac-3bc-3ab]=0\)

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

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

​Để \(N\)có nghĩa thì \(\left(a+b+c\right)^2\ne0\)

Hay \(a+b+c\ne0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

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

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall c,a\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Rightarrow a=b=c\)

Thay \(a=b=c\)vào \(N\), ta có : \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy \(N=\frac{1}{3}\)

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

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

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

Thay \(\hept{\begin{cases}x-y=-7\\xy=-6\end{cases}\left(3\right)}\)vào , ta được :

\(x^2+y^2=49-12=37\left(2\right)\)

Thay \(\left(2\right)\),\(\left(3\right)\)vào \(\left(1\right)\)vào , ta có giá trị của biểu thức tương đương với :

\(-7\left(37-6\right)-\left(-7^2\right)=-7.31-49=-266\)

a, \(\frac{5x}{x-1}+\frac{-5}{x-1}=\frac{5x-5}{x-1}=\frac{5\left(x-1\right)}{x-1}=5\)

b, \(\frac{1}{x-3}+\frac{2}{x+3}+\frac{9-x}{x^2-9}=\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{9-x}{\left(x-3\right)\left(x+3\right)}\)

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

Tương tự