Question

1) Tìm x, y, z biết x² + y² + z² + 3 = 2 (x + y + z).

2) Giải phương trình: \(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)

3) Tìm x, y thỏa mãn: 5x² + 5y² + 8xy + 2y - 2x + 2 = 0

Answer 1

1. \(x^2+y^2+z^2+3=2\left(x+y+z\right)< =>x^2-2x+1+y^2-2y+1+z^2-2z+1=0< =>\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)

=>x-1=0<=>x=1

y-1=0<=>y=1

z-1=0<=>z=1

vậy....

2. \(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)

<=>\(\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)

<=>\(\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)

<=>(2010-x)(1/2008-1/2009-1/2010)=0

vì 1/2008-1/2009-1/2010 khác 0 nên 2010-x=0<=>x=2010

Answer 2

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

\(\Leftrightarrow x^2-2x+1+y^2-2y+1+z^2-2z+1=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2=0\)

\(\Leftrightarrow x=y=z=1\)

2)\(\dfrac{2-x}{2008}-1=\dfrac{1-x}{2009}-\dfrac{x}{2010}\)

\(\Leftrightarrow\dfrac{2-x}{2008}+1=\dfrac{1-x}{2009}+1-\dfrac{x}{2010}+1\)

\(\Leftrightarrow\dfrac{2010-x}{2008}=\dfrac{2010-x}{2009}+\dfrac{2010-x}{2010}\)

\(\Leftrightarrow\left(2010-x\right)\left(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\right)=0\)

\(\Leftrightarrow x=2010\)(vì \(\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\ne0\))

Answer 3

3)\(5x^2+5y^2+8xy+2y-2x+2=0\)

\(\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow x=1;y=-1\)

Answer 4

\(5x^2+5y^2+8xy+2y-2x+2=0< =>\left(2x\right)^2+8xy+\left(2y\right)^2+x^2-2x+1+y^2+2y+1=0< =>\left(2x+2y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\)

<=> x-1=0<=>x=1

y+1=0<=>y=-1

2(x+y)=0<=>2(1+-1)=0