a,Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}CMR:\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
b, CHO \(\frac{b+c}{bc}=\frac{2}{a}CMR:\frac{b}{c}=\frac{a-b}{c-a}\)( cac ti so deu co nghia)
cho day ti so bang nhau \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
cmr\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
CMR:\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Từ\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}\Rightarrow4b^2z-6bcy=3acx-a^2z\)
\(\Rightarrow\)\(\left(4b^2+a^2\right)z=3c\left(ax+2by\right)\)
\(\Rightarrow\frac{z}{3c}=\frac{ax+2by}{a^2+4b^2}=\frac{ax}{a^2}=\frac{2by}{4b^2}\Rightarrow\frac{z}{3c}=\frac{x}{a}=\frac{y}{2b}\)
Với a,b,c khác 0, cho dãy tỉ số bằng nhau\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\).Chứng minh:\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
1 . Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
Chứng minh \(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
2 . Cho \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\left(\forall a,b,c\ne0;b\ne c\right)\). CMR : \(\frac{a}{b}=\frac{a-c}{c-b}\)
1.
Ta có : \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}\)
\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)
Áp dụng tính chất của dãy tỉ số bằng hau ta có :
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}\)
\(=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{cases}}\) \(\Rightarrow\hept{\begin{cases}2bz=3cy\\3cx=az\\ay=2bx\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{x}{3c}\left(đpcm\right)\)
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1. Sửa lại dòng cuối
\(\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Bài 2:
Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)
\(\Rightarrow2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\)
\(\Rightarrow ab-bc=ac-ab\)
\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Cho: \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
CMR: \(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
xem câu trả lời ở câu hỏi của bạn yến nhi nhé
Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\) . Chứng minh : \(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Ta có: \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}.\)
\(\Rightarrow\frac{a.\left(2bz-3cy\right)}{a^2}=\frac{2b.\left(3cx-az\right)}{4b^2}=\frac{3c.\left(ay-2bx\right)}{9c^2}.\)
\(\Rightarrow\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{2abz-3acy}{a^2}=\frac{6bcx-2abz}{4b^2}=\frac{3acy-6bcx}{9c^2}=\frac{2abz-3acy+6bcx-2abz+3acy-6bcx}{a^2+4b^2+9c^2}=\frac{\left(2abz-2abz\right)-\left(3acy-3acy\right)+\left(6bcx-6bcx\right)}{a^2+4b^2+9c^2}=0.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{2bz-3cy}{a}=0\\\frac{3cx-az}{2b}=0\\\frac{ay-2bx}{3c}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz-3cy=0\\3cx-az=0\\ay-2bx=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2bz=3cy\\3cx=az\\ay=2bx\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{z}{3c}=\frac{y}{2b}\\\frac{x}{a}=\frac{z}{3c}\\\frac{y}{2b}=\frac{x}{a}\end{matrix}\right.\Rightarrow\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\left(đpcm\right).\)
Chúc bạn học tốt!
\(Cho\)
\(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)
\(CMR:\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Giải hộ mình đanh cần gấp
Cho \(\frac{2bz-3cy}{a}=\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\).Chứng minh rằng : \(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)
Bạn không làm đc thì thui có ai khiến bạn trả lời lớp mấy đâu
Cho \(\frac{2bz-3cy}{a}\text{=}\frac{3cx-az}{2b}\text{=}\frac{ay-2bx}{3c}\)
Chứng minh : \(\frac{x}{a}\text{=}\frac{y}{2b}\text{=}\frac{z}{3c}\)