Cho dãy tỉ số: (bz-cy)/a=(cx-az)/b=(ay-bx)/c
CMR: x/a=y/b=z/c
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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 )
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
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cyx}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}=\frac{bxz-cxy+cyz-azy+ayz-bxz}{ax+by+cz}=0\)
\(\frac{bz-cy}{a}=0\Rightarrow bz=cy\Rightarrow\frac{b}{y}=\frac{c}{z}\)
\(\frac{cx-az}{b}=0\Rightarrow cx=az\Rightarrow\frac{c}{z}=\frac{a}{x}\)
\(\frac{ay-bx}{c}=0\Rightarrow ay=bx\Rightarrow\frac{a}{x}=\frac{b}{y}\)
\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\left(đpcm\right)\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Rightarrow\frac{bxz-cxy}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}=\frac{bxz-cxy+cxy-azy+ayz-bxz}{ax+by+cz}=\frac{0}{ax+by+cz}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}\Rightarrow}\hept{\begin{cases}\frac{z}{c}=\frac{y}{b}\left(1\right)\\\frac{x}{a}=\frac{z}{c}\left(2\right)\\\frac{y}{b}=\frac{x}{a}\left(3\right)\end{cases}}}\)
Từ (1),(2),(3) suy ra \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)