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Link : Bảy hằng đẳng thức đáng nhớ – Wikipedia tiếng Việt
Dùng hằng mở rộng số 4
Ta có :
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\) (1)
Lại có :
\(\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)^2=\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1^2=1\) (chỗ này dùng cái skill mở rộng)
<=> \(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\left(\frac{xyc}{abc}+\frac{ayz}{abc}+\frac{bzx}{abc}\right)=1\)
<=> \(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}+2.\frac{ayz+bxz+cxy}{abc}=1\)
Thay 1 vào
=> \(\frac{a^2}{x^2}+\frac{b^2}{y^2}+\frac{c^2}{z^2}=1\)
Ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\cdot\frac{xyc+yza+zxb}{abc}=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Leftrightarrow\frac{yza+zxb+xyc}{xyz}=0\)
\(\Rightarrow yza+zxb+xyc=0\)
\(\Rightarrow A=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
Ta có : \(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)
Suy ra : \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-\frac{x^2}{a^2+b^2+c^2}-\frac{y^2}{a^2+b^2+c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\frac{x^2}{a^2}-\frac{x^2}{a^2+b^2+c^2}+\frac{y^2}{b^2}-\frac{y^2}{a^2+b^2+c^2}+\frac{z^2}{c^2}-\frac{z^2}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)+y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)+z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
Vì : \(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right);y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right);z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)\ge0\forall x\)
Nên : \(x^2\left(\frac{1}{a^2}-\frac{1}{a^2+b^2+c^2}\right)=0;y^2\left(\frac{1}{b^2}-\frac{1}{a^2+b^2+c^2}\right)=0;z^2\left(\frac{1}{c^2}-\frac{1}{a^2+b^2+c^2}\right)=0\)
=> x = 0 ; y = 0 ; z = 0
Vậy x + y + z = 0 (đpcm)
câu 1 là :từ a/x + b/y + c/z =0 suy ra (ayz+bxz+cxy)/xyz =0 suy ra ayz+bxz+cxy=0 (1)
vì x/a + y/b + z/c =1 (gt) suy ra (x/a + y/b + z/c )^2 = 1^2 . suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2(xy/ab + yz/bc + xz/ac) =1
suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 + 2[(ayz+bxz+cxy)/abc = 1 (2)
Từ (1) và (2) suy ra x^2/a^2 + y^2/b^2 + z^2/c^2 =1 (đpcm)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{cxy+ayz+bxz}{abc}\right)=1\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{0}{abc}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.0=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(dpcm\right)\)
Chúc bạn học tốt
1 cái T I C K nha cảm ơn
chịu khó lắm
Ok
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\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow\frac{ayz+bxz+cxy}{xyz}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{xbc+yac+zab}{abc}=1\)
\(\Rightarrow xbc+yac+zab=abc\)
\(\Rightarrow\left(xbc\right)^2+\left(yac\right)^2+\left(zab\right)^2+2.xbc.yac+2.yac.zab+2.xbc.zab=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2+2abc\left(cxy+ayz+bxz\right)=\left(abc\right)^2\)
\(\Rightarrow x^2b^2c^2+y^2a^2c^2+z^2a^2b^2=a^2b^2c^2\)
\(\Rightarrow\frac{x^2b^2c^2+y^2a^2c^2+z^2a^2b^2}{a^2b^2c^2}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Rightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\)
\(\left(\frac{x}{a}+\frac{y}{b}\right)^2+2\left(\frac{x}{a}+\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\left(\frac{x}{a}\right)^2+2\frac{x}{a}\frac{y}{b}+\left(\frac{y}{b}\right)^2+\left(2\frac{x}{a}+2\frac{y}{b}\right)\frac{z}{c}+\left(\frac{z}{c}\right)^2=1\)
\(\frac{x^2}{a^2}+\frac{2xy}{ab}+\frac{y^2}{b^2}+\frac{2xz}{ac}+\frac{2yz}{bc}+\frac{z^2}{c^2}=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\left(\frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{c}{z}+\frac{b}{y}+\frac{a}{x}\right)=1\)
\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)
\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(ĐPCM\right)\)
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)\)
\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)