Cho \(\frac{cy-bz}{x}\)= \(\frac{az-cx}{y}\)=\(\frac{bx-ay}{z}\)
Chứng minh rằng: \(\frac{a}{x}\)=\(\frac{b}{y}\)=\(\frac{c}{z}\)
Biết:
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\). Chứng minh rằng:
\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
Ta có :
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\Rightarrow\frac{a\left(bz-cy\right)}{a^2}=\frac{b\left(cx-az\right)}{b^2}=\frac{c\left(ay-bx\right)}{c^2}\)
\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcz-abz}{b^2}=\frac{acy-bcz}{c^2}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{abz-acy+bcx-abz+acy-bcz}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0\)
=> abz - acy = 0 => abz = acy => bz = cy (1)
bcx - abz = 0 => bcx = abz => cx = az (2)
acy - bcx = 0 => acy = bcx => ay = bx
Chuyển đổi vế 1 và 2 ta có :
\(bz=cy\Rightarrow\frac{b}{y}=\frac{c}{z}\left(a\right)\)
\(cx=az\Rightarrow\frac{c}{z}=\frac{a}{x}\left(b\right)\)
Từ a và b
=> \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\) (ĐPCM)
Cho dãy tỉ số \(\frac{bz-cy}{a}=\frac{cx-az}{z}=\frac{ay-bx}{c}\). Chứng minh rằng :\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
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 )
Cho a,b,c là ba số không âm thỏa mãn \(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)
Chứng minh rằng:\(\left(ax+by+cz\right)^2=\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)\)
\(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}\)
\(\Rightarrow\frac{acy-bcx}{c^2}=\frac{bcx-abz}{b^2}=\frac{abz-acy}{a^2}=\frac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}}\)
\(\Rightarrow\left(ay-bx\right)^2+\left(cx-az\right)^2+\left(bz-ay\right)^2=0\)
\(\Rightarrow a^2y^2-2axby+b^2x^2+a^2z^2-2axcz+c^2x^2+b^2z^2-2bycz\)
\(+c^2y^2=0\)
\(\Rightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)
\(=a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz\)
\(\Rightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)
\(Cho:\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}.CMR:\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)
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: \(\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}^{\left(đpcm\right)}}\)
Cho \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\). Chứng minh rằng \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
#)Tuy k giải được nhưng có bài cho tham khảo nek :
Câu hỏi của Hann Hann - Toán lớp 7 - Học toán với OnlineMath
Link : https://olm.vn/hoi-dap/detail/7941323649.html
Mk sẽ gửi về chat cho
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 : \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)
\(\Rightarrow bz-cy=cx-az=ay-bx=0\)
\(\Rightarrow\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=0\left(ĐPCM\right)\)
cho \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\) chứng minh \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
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
biết rằng:\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\).hãy chứng minh :\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
áp dụng tính chất hai dãy tỉ số bằng nhau nha bạn
Cho \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\). CMR: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Cho \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\) chứng minh rằng: \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)
\(\Rightarrow\hept{\begin{cases}x=ak\\y=bk\\z=ck\end{cases}}\)
Ta có \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Rightarrow\frac{bck-cbk}{a}=\frac{cak-ack}{b}=\frac{abk-bak}{c}\)
\(\Rightarrow\frac{0}{a}=\frac{0}{b}=\frac{0}{c}\)
\(\Rightarrow0=0=0\)(đpcm)
\(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\Rightarrow\hept{\begin{cases}bx=ay\\cx=az\\cy=bz\end{cases}\Rightarrow\hept{\begin{cases}ay-bx=0\\cx-az=0\\bz-cy=0\end{cases}\Rightarrow}}\)\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=0\left(đpcm\right)\)