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

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

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

=> 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 )

Câu hỏi của GT 6916 - Toán lớp 7 - Học toán với OnlineMath tham khảo

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

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

Áp dụng t/c ãy tỉ số bằng nhau, ta có:

a) Ta có: A = ax + bx + cx + ay + by + cy + az + bz + cz

                  = x.(a+b+c) + y.(a+b+c) + z.(a+b+c)

                  = (a+b+c).(x+y+z) (1)

Lại có: a + b + c = -3 (2)

            x + y + z = -6 (3)

Từ (1) ; (2) ; (3) => A = -3.(-6) = 18

           Vậy A = 18

b) B = ax - bx - cx - ay + by + cy - az + bz +cz

       = x.(a-b-c) - y.(a-b-c) - z.(a-b-c)

       = (a-b-c).(x-y-z)

Lại có: a - b - c = 0 ; x - y - z = 2016

=> B = 0.2016 = 0

Vậy B = 0

Ta có: bx−cyabx−cya = cx−axbcx−azb = ay−bxcay−bxc

⇒ bx−cyabx−cya = a(bx−cy)a²a(bx−cy)a²  = abx−acya²abx-acya²

    cx−azbcx−axb = b(cx−az)b²b(cx−az)b² = bcx−baxb²bcx−baxb²

     ay−bxcay−bxc = c(ay−bx)c²c(ay−bx)c² = cay−cbxc²cay−cbxc²

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

bx−cyabx−cya = cx−azbcx−axb = cy−bxccy−bxc = abx−acy+bcx−bax+cay−cbxa²+b²+c²abx−acy+bcx−bax+cay−cbxa²+b²+c²  = 0

\(\Rightarrow\) bx - cy = 0

    cx - ax = 0

    ay - bx = 0

\(\Rightarrow\) bx = cy

    cx = ax

    ay = bx

\(\Rightarrow\) xcxc = ybyb 

    xaxa = xcxc 

    ybyb = xaxa 

\(\Rightarrow\) xaxa = ybyb = xcxc 

cyabx o dau vay

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cyx}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}=\frac{bxz-cxy+cyz-azy+ayz-bxz}{ax+by+cz}=0\)

\(\frac{bz-cy}{a}=0\Rightarrow bz=cy\Rightarrow\frac{b}{y}=\frac{c}{z}\)

\(\frac{cx-az}{b}=0\Rightarrow cx=az\Rightarrow\frac{c}{z}=\frac{a}{x}\)

\(\frac{ay-bx}{c}=0\Rightarrow ay=bx\Rightarrow\frac{a}{x}=\frac{b}{y}\)

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

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

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

\(\Rightarrow\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{z}{c}=\frac{y}{b}\left(1\right)\\\frac{x}{a}=\frac{z}{c}\left(2\right)\\\frac{y}{b}=\frac{x}{a}\left(3\right)\end{cases}}}\)

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