Bài 2:

a) \(x^2+4x-y^2+4=\left(x^2+2\cdot x\cdot2+2^2\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)

b) \(3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)

c) \(x^2-2xy+y^2-z^2+2zt-t^2=\left(x-y\right)^2-\left(z-t\right)^2=\left(x-y-z+t\right)\left(x-y+z-t\right)\)

1)

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

b) \(xz+yz-5\left(x+y\right)=z\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(z-5\right)\)

c) \(3x^2-3xy-5x+5y=\left(3x^2-3xy\right)-\left(5x-5y\right)=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)

\(\sin45-cotg60\cdot\cos30=\dfrac{\sqrt{2}}{2}+\dfrac{1}{\tan\left(60\right)}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}=\dfrac{1+\sqrt{2}}{2}\)

p/s: cot60 = \(\dfrac{1}{\tan60}\)

Bài 3:

a) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\left(2x+y\right)^2:\left(2x+y\right)=2x+y\)

b) \(\left(27x^3+1\right):\left(3x+1\right)=\left(3x+1\right)\left(9x^2-9x+1\right):\left(3x+1\right)=9x^2-9x+1\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\left(x-3y\right)^2:\left(3y-x\right)=\left(3y-x\right)^2:\left(3y-x\right)=3y-x\)

d) \(\left(8x^3-1\right):\left(4x^2+2x+1\right)=\left(2x-1\right)\left(4x^2+2x+1\right):\left(4x^2+2x+1\right)=2x-1\)

Bài 4: Tương tự bài 3 '-'

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\(x^2-2x+y^2-4y+7\)

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

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

Có: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(y-2\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(\Rightarrow A_{Min}=2\) xảy ra khi

\(\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(x^3+7y=y^3+7x\)

\(\Leftrightarrow x^3-y^3-7x+7y=0=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-7\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\Rightarrow x=y\left(loai\right)\\x^2+xy+y^2-7=0\left(1\right)\end{matrix}\right.\)

pt (1)<=> \(x^2+xy+y^2=7\) (*)

Giải (*) ta đc nghiệm phân biệt:

x = 1 và y = 2

hoặc x = 2 ; y = 1

*) \(x^2-6x+9=x^2-2\cdot x\cdot3+3^2=\left(x-3\right)^2\)

*) \(4x^2-36=\left(2x\right)^2-6^2=\left(2x-6\right)\left(2x+6\right)\)

*) \(8-x^3=2^3-x^3=\left(2-x\right)\left(4+2x+x^2\right)\)