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1) \(A=\left(x+y\right)^2+4xy=x^2+2xy+y^2+4xy=x^2+6xy+y^2\)
2) \(B=\left(6x-2\right)^2+4\left(3x-1\right)\left(2+y\right)+\left(y+2\right)^2\)
\(=\left(6x-2\right)^2+2\left(6x-2\right)\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(6x-2+y+2\right)^2=\left(6x+y\right)^2=36x^2+12xy+y^2\)
3) \(C=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)
\(=\left(x-y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x-y+x+y\right)^2=\left(2x\right)^2=4x^2\)
d: \(\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x+3\right)\left(x-3\right)}\)
\(\dfrac{1}{3-x}=\dfrac{-1}{x-3}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x^2-9}=\dfrac{1}{\left(x+3\right)\left(x-3\right)}\)
Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
a) MTC: \(12x^3y^3\)
\(\dfrac{3}{4x^3y^2}=\dfrac{3\cdot3y}{4x^3y^2\cdot3y}=\dfrac{9y}{12x^3y^3}\)
\(\dfrac{2}{3xy^3}=\dfrac{2\cdot4x^2}{3xy^3\cdot4x^2}=\dfrac{8x^2}{12x^3y^3}\)
b) MTC: \(x\left(x-3\right)^2\)
\(\dfrac{5}{x^2-6x+9}=\dfrac{5}{\left(x-3\right)^2}=\dfrac{5x}{x\left(x-3\right)^2}\)
\(\dfrac{3}{x^2-3x}=\dfrac{3}{x\left(x-3\right)}=\dfrac{3\left(x-3\right)}{x\left(x-3\right)^2}=\dfrac{3x-9}{x\left(x-3\right)^2}\)
