Vì \(\frac{a-x}{b-y}=\frac{a}{b}\) nên \(\left(a-x\right).b=\left(b-y\right).a\) ;   \(ab-xb=ba-ya\)

Do đó : \(xb=ya\)       hay \(\frac{x}{y}=\frac{a}{b}\)(đpcm)

Vậy ___________________________

Vì \(\frac{a-x}{b-y}=\frac{a}{b}\) nên (a - x) . b = (b - y) . a

\(\Leftrightarrow\) ab - xb = ba - ya

Do ab = ba \(\Rightarrow\) xb = ya hay \(\frac{x}{y}=\frac{a}{b}\)

     ta co :\(\frac{a-x}{b-y}=\frac{a}{b}\Rightarrow b\left(a-x\right)=a\left(b-y\right)\)

                                          \(\Rightarrow ba-bx=ab-ay\)

                                           \(\Rightarrow ba+ay=bx+ab\)

                                           \(\Rightarrow ay=bx\)

                                           \(\Rightarrow\frac{x}{y}=\frac{a}{b}\)

Minh chac chan 100% tick cho minh nha

Theo giả thiết suy ra \(\frac{a\left(y+z\right)}{abc}=\frac{b\left(z+x\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)\(\Rightarrow\)\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{z+x-\left(y+z\right)}{ac-bc}=\frac{x-y}{c\left(a-b\right)}\) (1)

\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{y+z-\left(x+y\right)}{bc-ab}=\frac{z-x}{b\left(c-a\right)}\) (2)

\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{x+y-\left(z+x\right)}{ab-ac}=\frac{y-z}{a\left(b-c\right)}\) (3)

Từ (1), (2), (3) suy ra \(\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\) (đpcm).

(đpcm) Tức là : đá phải con mèo

(đpcm)tức là điều phải chứng minh

Có: 1+x = \(\frac{a+b+a-b}{a+b}\) = \(\frac{2a}{a+b}\)

Tương tự, 1 + y = \(\frac{2b}{b+c}\)

1 + z = \(\frac{2c}{c+a}\)

1 - x = \(\frac{q+b-a+b}{a+b}\) = \(\frac{2a}{a+b}\)

Tương tự như thế rồi nhân (1+x), (1+y), (1+z) với nhau; (1-z), (1-y), (1-z) với nhau

1, 

\(\frac{a+2}{a-2}=\frac{b+3}{b-3}\)

<=> (a - 2)(b + 3) = (a + 2)(b - 3)

<=> ab + 3a - 2b - 6 = ab - 3a + 2b - 6

<=> 3a - 2b = -3a + 2b

<=> 6a = 4b

<=> 3a = 2b 

<=> \(\frac{a}{2}=\frac{b}{3}\)(Đpcm)

2,

Có:

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(=\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)

\(=\frac{abz-acy+bcx-baz+cay-cbx}{a^2+b^2+c^2}=0\)

=> bz - cy = 0

=> bz = cy

=> \(\frac{b}{y}=\frac{c}{z}\)(1)

=> cx - az = 0

=> cx = az

=> \(\frac{c}{z}=\frac{a}{x}\)(2)

Từ (1) và (2)

=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)(Đpcm)

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

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

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

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

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

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

\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}\left(\frac{c}{z}+\frac{b}{y}+\frac{a}{x}\right)=1\)

\(\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right)+\frac{2xyz}{abc}.0=1\)

\(\Rightarrow\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\left(ĐPCM\right)\)

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

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

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

\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)