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3:
\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{\left(a+b\right)-\left(a-b\right)}{\left(c+d\right)-\left(c-d\right)}=\dfrac{2b}{2d}=\dfrac{b}{d}\) (1)
\(\Rightarrow\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}=\dfrac{a+b+a-b}{c+d+c-d}=\dfrac{2a}{2c}=\dfrac{a}{c}\) (2)
Từ (1) và (2) suy ra \(\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\) \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\) ĐPCM
Ta có:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)
\(\Leftrightarrow\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\)
\(\Leftrightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(đpcm\right)\)
ta có:
a² + b² + c² = (a + b + c)² - 2(ab + bc + ca). (*)
rất dể cm: ta khai triển hằng đẳng thức (a + b + c)² rồi rút gọn là ra (*)
AD (*):
x²/a² + y²/b² + z²/c² = (x/a + y/b + z/c)² - 2(xy/ab + yz/bc + zx/ca) =
= 1² - 2(xyz/abc)(c/z + a/x + b/y) = 1 + 0 = 1
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\), suy ra \(\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1\)
\(\Leftrightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(\Leftrightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\frac{xyz}{abc}\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)=1\) \(\left(1\right)\)
Mà \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\) (theo gt) nên từ \(\left(1\right)\) \(\Rightarrow\) \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\) \(\left(đpcm\right)\)
Vi de bai bao c/m
Nen se dung la nhu the
Neu khong dung nhu the
Chung to de bai... SAI !!!
He he...
Ta có:
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\left(1\right)\)
Mặt khác:
\(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ac}\right)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2.\frac{cxy+ayz+bzx}{abc}=1\left(2\right)\)
Từ (1) và (2) ta có đpcm.
\(1+1=2\)
1+1=2
là sai