\(\frac{cy-bz}{x}=\frac{az-cx}{y}=\frac{bx-ay}{z}\)

=> \(\frac{cyx-bzx}{x^2}=\frac{azy-cxy}{y^2}=\frac{bxz-ayz}{z^2}=\frac{cyx-bzx+azy-cxy+bzx-ayz}{x^2+y^2+z^2}\)

                                                                                    \(=\frac{0}{x^2+y^2+z^2}=0\)

Khi đó \(\hept{\begin{cases}cyx-bzx=0\\azy-cxy=0\\bxz-ayz=0\end{cases}}\Leftrightarrow\hept{\begin{cases}cy=bz\\az=cx\\bx=ay\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{b}{y}=\frac{c}{z}\\\frac{c}{z}=\frac{a}{x}\\\frac{z}{x}=\frac{b}{y}\end{cases}}\Leftrightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)

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

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

\(\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}}\)

\(\Rightarrow\left(ay-bx\right)^2+\left(cx-az\right)^2+\left(bz-ay\right)^2=0\)

\(\Rightarrow a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz\)

\(+c^2y^2=0\)

\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)

\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

1/ 

Từ \(a-b=2\left(a+b\right)\Rightarrow a-b=2a+2b\Rightarrow a-2a=2b+b\Rightarrow-a=3b\Rightarrow a=-3b\)

\(\Rightarrow\frac{a}{b}=\frac{-3b}{b}=-3\)

\(\Rightarrow\hept{\begin{cases}a-b=-3\\2\left(a+b\right)=-3\end{cases}\Rightarrow\hept{\begin{cases}a-b=-3\\a+b=-\frac{3}{2}\end{cases}}}\)

\(\Rightarrow a-b+a+b=-3-\frac{3}{2}\Rightarrow2a=\frac{-9}{2}\Rightarrow a=\frac{-9}{4}\)

Có: \(a-b=-3\Rightarrow b=a+3\Rightarrow b=\frac{-9}{4}+3=\frac{3}{4}\)

Vậy a=-9/4,b=3/4

2/ Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\Rightarrow x=ak,y=bk,z=ck\)

Ta có: \(\frac{bx-ay}{a}=\frac{bak-abk}{a}=0\left(1\right)\)

\(\frac{cx-az}{y}=\frac{cak-ack}{y}=0\left(2\right)\)

\(\frac{ay-bx}{c}=\frac{abk-bak}{c}=0\left(3\right)\)

Từ (1),(2),(3) => đpcm

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

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

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

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

\(\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}}\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{x}{a}=\frac{z}{c}\\\frac{y}{b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

* C1 :(bz - cy)/a = (abz - acy)/a²

(cx - az)/b = (bcx - abz)/b²

(ay - bx)/c = (acy - bcx)/c²

Mà (bz - cy)/a = (cx - az)/b = (ay - bx)/c

=>(abz - acy)/a² = (bcx - abz)/b² = (acy - bcx)/c² = (abz - acy + bcx - abz + acy - bcx)/a² + b² + c² = 0

=>(bz - cy)/a = (cx - az)/b = (ay - bx)/c = 0

=>bz - cy = cx - az = ay - bx = 0

*Xét bz - cy = 0

=>bz = cy

=>z/c = y/b

Chứng minh tương tự = >x/a = y/b ; x/a = z/c

=> x/a = y/b = z/c

*C2 : 

(bz - cy)/a = (abz - acy)/ax

(cx - az)/by = (bcx - abz)/by

(ay - bx)/cz = (acy - bcx)/cz

Làm tương tự như C1

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

Suy ra: \(\frac{bxz-cxy}{ax}=\frac{cxy-azy}{bx}=\frac{azy-bxz}{cx}\). Áp dụng tính chất dãy tỉ số bằng nhau có: 

\(\frac{bxz-cxy}{ax}=\frac{cxy-azy}{bx}=\frac{azy-bxz}{cx}=\frac{\left(bxz-cxy\right)+\left(cxy-azy\right)+\left(azy-bxz\right)}{ax+bx+cx}\)

\(=\frac{\left(bxz-bxz\right)-\left(cxy-cxy\right)-\left(azy-azy\right)}{ax+by+cz}=\frac{0}{ax+by+cz}\)

Suy ra: \(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Leftrightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)

Áp dụng tính chất tỉ lệ thức ta được: \(\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}^{\left(đpcm\right)}}\)

áp dụng tính chất hai dãy tỉ số bằng nhau nha bạn

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Giải:

Đặt : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)  => \(\hept{\begin{cases}x=ak\\y=bk\\z=ck\end{cases}}\)

Khi đó, ta có:

\(\frac{b.ck-c.bk}{a}=\frac{0}{a}=0\) (1)

\(\frac{c.ak-a.ck}{b}=\frac{0}{b}=0\) (2)

\(\frac{a.bk-b.ak}{c}=\frac{0}{c}=0\) (3)

Từ (1); (2); (3) suy ra \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

Ta có : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)

\(\Rightarrow bz-cy=cx-az=ay-bx=0\)

\(\Rightarrow\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=0\left(ĐPCM\right)\)

mk k viết đề nha bạn!

\(=>\frac{a\left(bz-cy\right)}{a^2}=\frac{b\left(cx-az\right)}{b^2}=\frac{c.\left(by-ax\right)}{c^2}\)

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

\(=>\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bc}{c}=0\)

=> bz - cy = cx - az = ay - bx = 0

+) bz - cy = 0 => bz = cy => y / b = z/c 

+) cx - az = 0 => cx = az => x / a = z/ c
=> x / a = y / b = z/ c ( dpcm )

vi bz-cy/a=cx-az/b=ay-bx/c=>a(bz-cy)/a^2=b(cx-az)/b^2=c(ay-bx)/c^2

=>abz-acy/a^2=bcx-abz/b^2=cay-cbx/c^2=>abz-acy+bcx-abz+cay-cbx/a^2+b^2+c^2

=>o/a^2+b^2+c^2=0

=>bz-cy=0=>y/b=z/c(1)

cx-az=o=>x/a=z/c(2)

từ (1) và (2) =>x/a=y/b=z/c

Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\Leftrightarrow\frac{baz-cay}{a^2}=\frac{cbx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{baz-cay+cbx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow bz=cy\Leftrightarrow\frac{y}{b}=\frac{z}{c}\)

\(\Rightarrow cx=az\Leftrightarrow\frac{x}{a}=\frac{z}{c}\) 

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

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