cho : x/a + y/b + z/c =1 và a/x +b/y + c/z =0 hãy tính: (x^2 /a^2) +( y^2/b^2) +( z^2 + c^2) 17/09/2021 Bởi Parker cho : x/a + y/b + z/c =1 và a/x +b/y + c/z =0 hãy tính: (x^2 /a^2) +( y^2/b^2) +( z^2 + c^2)
Giải thích các bước giải: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \rightarrow (\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c})^{2}=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ca})=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2\dfrac{xyz}{abc}(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z})=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2\dfrac{xyz}{abc}.0=1\\ \rightarrow \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1\\$ Bình luận
Ta có: `a/x+b/y+c/z=0` `⇒ayz+bxz+cxy=0` Ta có: ` x/a+y/b+z/c=1` `⇔(x/a+y/b+z/c)^2=1` `⇔x^2/a^2+y^2/b^2+z^2/c^2+2((xy)/(ab)+(yz)/(bc)+(zx)/(ca))=1` `⇔x^2/a^2+y^2/b^2+z^2/c^2+2((ayz+bxz+cxy)/(abc))=1` `⇔x^2/a^2+y^2/b^2+z^2/c^2+2.0=1` `⇔x^2/a^2+y^2/b^2+z^2/c^2=1` `⇒ĐPCM` Bình luận
Giải thích các bước giải: $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \rightarrow (\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c})^{2}=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ca})=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2\dfrac{xyz}{abc}(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z})=1\\ \rightarrow (\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}})+2\dfrac{xyz}{abc}.0=1\\ \rightarrow \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}+\dfrac{z^{2}}{c^{2}}=1\\$
Ta có:
`a/x+b/y+c/z=0`
`⇒ayz+bxz+cxy=0`
Ta có:
` x/a+y/b+z/c=1`
`⇔(x/a+y/b+z/c)^2=1`
`⇔x^2/a^2+y^2/b^2+z^2/c^2+2((xy)/(ab)+(yz)/(bc)+(zx)/(ca))=1`
`⇔x^2/a^2+y^2/b^2+z^2/c^2+2((ayz+bxz+cxy)/(abc))=1`
`⇔x^2/a^2+y^2/b^2+z^2/c^2+2.0=1`
`⇔x^2/a^2+y^2/b^2+z^2/c^2=1`
`⇒ĐPCM`