Question

cho x, y, z thỏa mãn (\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)):(\(\dfrac{1}{x+y+z}\))=1

tính giá trị biểu thức B=(x²¹+y²¹)(y¹¹+z¹¹)(z²⁰¹⁷+x²⁰¹⁷)

Answer 1

Lời giải:

Ta có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)

\(\Leftrightarrow \frac{xy+yz+xz}{xyz}.(x+y+z)=1\Leftrightarrow (xy+yz+xz)(x+y+z)=xyz\)

\(\Leftrightarrow xy(x+y)+yz(y+z)+xz(x+z)+2xyz=0\)

\(\Leftrightarrow xy(x+y+z)+yz(y+z+x)+xz(x+z)=0\)

\(\Leftrightarrow y(x+y+z)(x+z)+xz(x+z)=0\)

\(\Leftrightarrow (x+z)[y(x+y+z)+xz]=0\)

\(\Leftrightarrow (x+z)(y+x)(y+z)=0\)

Do đó:

\(B=(x+y)(x^{20}+....+y^{20})(y+z)(y^{10}+...+z^{10})(z+x)(z^{2016}+x^{2016})\)

\(=(x+y)(y+z)(x+z)(x^{20}+..+y^{20})(y^{10}+..+z^{10})(z^{2016}+x^{2016})=0\)

Answer 2

Lời giải:

Ta có:
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right):\left(\frac{1}{x+y+z}\right)=1\)

\(\Leftrightarrow \frac{xy+yz+xz}{xyz}.(x+y+z)=1\Leftrightarrow (xy+yz+xz)(x+y+z)=xyz\)

\(\Leftrightarrow xy(x+y)+yz(y+z)+xz(x+z)+2xyz=0\)

\(\Leftrightarrow xy(x+y+z)+yz(y+z+x)+xz(x+z)=0\)

\(\Leftrightarrow y(x+y+z)(x+z)+xz(x+z)=0\)

\(\Leftrightarrow (x+z)[y(x+y+z)+xz]=0\)

\(\Leftrightarrow (x+z)(y+x)(y+z)=0\)

Do đó:

\(B=(x+y)(x^{20}+....+y^{20})(y+z)(y^{10}+...+z^{10})(z+x)(z^{2016}+x^{2016})\)

\(=(x+y)(y+z)(x+z)(x^{20}+..+y^{20})(y^{10}+..+z^{10})(z^{2016}+x^{2016})=0\)