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24 tháng 3 2017

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=\dfrac{xbc+yac+zab}{abc}=1\\ \Rightarrow xbc+yac+zab=abc\)

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=\dfrac{ayz+bxz+cxy}{xyz}=0\\ \Rightarrow ayz+bxz+cxy=0\)

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(abc\right)^2}\)

\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc+yac+zab\right)^2}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc\left(ayz+bxz+cxy\right)}\)

\(\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2abc.0}\\ =\dfrac{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}{\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2}=1\)

vậy \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)(đpcm)

25 tháng 3 2017

\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.\left(\dfrac{c}{z}+\dfrac{b}{y}+\dfrac{a}{x}\right)=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2.\dfrac{xyz}{abc}.0=1\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)

13 tháng 6 2017

* Ta có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

\(\Leftrightarrow\dfrac{axy}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)

\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)

\(\Leftrightarrow ayz+bxz+cxy=0\)

* Ta có:

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{b^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)\(cxy+bxz+ayz=0\)

\(\Rightarrow2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=0\)

\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

Vậy.........................

13 tháng 6 2017

Ta có:

\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

=>\(\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)

=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)

=>\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{ayz}{abc}+\dfrac{bxz}{abc}\right)=1\) (1)

Lại có:

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)

=> \(\dfrac{a}{x}.\dfrac{yz}{yz}+\dfrac{b}{y}.\dfrac{xz}{xz}+\dfrac{c}{z}.\dfrac{xy}{xy}=0\)

=>\(\dfrac{ayz}{xuy}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\) (2)

Thay (2) vào (1) ta được

\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)

=> \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

2 tháng 3 2017

vi a/x + b/y + c/z =0 suy ra ayz/xyz + bxz/xyz + cxy/xyz =0 suy ra ayz+bxz+cxy /xyz =0 suy ra ayz + bxz + cxy =0

vi x/a + y/b =z/c =0 suy ra (x/a + y/b + z/c )^2 =0 suy ra x^2/a^2 +y^2/b^2 + z^2/c^2 + 2(xy/ab + xz/ac + yz/bc) =0

suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(cxy+ bxz +ayz /abc) =0

suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =0

suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 +2011 = 2011

10 tháng 12 2017

Ta có : \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\)

\(\Rightarrow\dfrac{bcx}{abc}+\dfrac{acy}{abc}+\dfrac{abz}{abc}=0\)

\(\Rightarrow\dfrac{bcx+acy+abz}{abc}=0\)

\(\Rightarrow bcx+acy+abz=0\)

\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=10\)

\(\Rightarrow\left(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\right)^2=10^2\)

\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{ab}{xy}+\dfrac{bc}{yz}+\dfrac{ac}{xz}\right)=100\)

\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{abz+bcx+acy}{xyz}\right)=100\)

\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}+2\left(\dfrac{0}{xyz}\right)=100\)

\(\Rightarrow\dfrac{a^2}{x^2}+\dfrac{b^2}{y^2}+\dfrac{c^2}{z^2}=100\)

\(\Rightarrow S=100\)

10 tháng 12 2017

Ta đặt x/a=m;y/b=n;c/z=k suy ra m+n+k=0 suy ra

a/x=1/m;y/b=1/n và c/z=1/k từ đó ta có bài toán m+n+k=0 và 1/n+1/m+1/k=1

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