Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Rightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\Rightarrow\left(10a+b\right)\left(b+c\right)=\left(10b+c\right)\left(a+b\right)\)
\(\Rightarrow10ab+b^2+10ac+bc=10ab+ac+10b^2+bc\Rightarrow9b^2=9ac\Rightarrow b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(đpcm\right)\)
Ta có:
\(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}\)
Mà: \(\left\{\begin{matrix}\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{10a+b+10b+c}{a+b}=9a+10b+c\\\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{10b+c+10c+a}{b+c}=9b+10c+a\\\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{10c+a+10a+b}{c+a}=9c+10a+b\end{matrix}\right.\)
\(\Rightarrow9a+10b+c=9b+10c+a=9c+10a+b\)
\(\Rightarrow\left\{\begin{matrix}9a=9b=9c\\10b=10c=10a\\c=a=b\end{matrix}\right.\)\(\Rightarrow a=b=c\)
Vậy \(a=b=c\) (Đpcm)
2) Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{ab}{b}=\frac{bc}{c}=\frac{ca}{a}=\frac{ab+bc+ca}{b+c+a}=\frac{\left(10a+b\right)+\left(10b+c\right)+\left(10c+a\right)}{a+b+c}=\frac{11.\left(a+b+c\right)}{a+b+c}=11\)
\(\Rightarrow\begin{cases}ab=11b\\bc=11c\\ca=11a\end{cases}\)\(\Rightarrow\begin{cases}10a+b=11b\\10b+c=11c\\10c+a=11a\end{cases}\)\(\Rightarrow\begin{cases}10a=10b\\10b=10c\\10c=10a\end{cases}\)\(\Rightarrow10a=10b=10c\)
=> a = b = c (đpcm)
soyeon_Tiểubàng giải bạn giúp bn ấy ik trong đó có câu 2 mk cần ó
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}-\overline{bc}-\overline{ca}+\overline{ca}+\overline{ab}}{a+b-b-c+c+a}=\frac{2\overline{ab}}{2a}=10+\frac{b}{a}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{\overline{ab}+\overline{bc}+\overline{bc}+\overline{ca}-\overline{ca}-\overline{ab}}{a+b+b+c-c-a}=\frac{2\overline{bc}}{2b}=10+\frac{c}{b}\)
+ \(\frac{\overline{ab}+\overline{bc}}{a+b}=\frac{\overline{bc}+\overline{ca}}{b+c}=\frac{\overline{ca}+\overline{ab}}{c+a}=\frac{-\overline{ab}-\overline{bc}+\overline{bc}+\overline{ca}+\overline{ca}+\overline{ab}}{-a-b+b+c+c+a}=\frac{2\overline{ca}}{2c}=10+\frac{a}{c}\)
=> \(\frac{b}{a}=\frac{c}{b}=\frac{a}{c}\Rightarrow\frac{b+c+a}{a+b+c}=1\Rightarrow a=b=c\)
Có : \(\dfrac{\overline{ab}}{\overline{bc}}=\dfrac{a}{c}\Rightarrow\dfrac{10a+b}{10b+c}=\dfrac{a}{c}=\dfrac{9a+b}{10b}\)( áp dụng dãy tỉ số bằng nhau)
\(=\dfrac{111...11.\left(9a+b\right)}{111..11.10b}\)(có n chữ số 1 trong số 111..111)
\(\dfrac{999..99a+111..11b}{111..110b}=\dfrac{a}{c}=\dfrac{999..99a+a+111..11b}{111..110b+c}=\dfrac{100...000a+111...11b}{111..110b+c}\)=\(\dfrac{\overline{abbb...bb}}{\overline{bbb..bbc}}=\dfrac{a}{c}\)
Ta có:
\(\frac{\overline{ab}}{\overline{bc}}=\frac{b}{c}\)
<=> \(\frac{a.10+b}{b.10+c}=\frac{b}{c}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{a.10+b}{b.10+c}=\frac{b}{c}=\frac{10a+b-b}{10b+c-c}=\frac{10a}{10b}=\frac{a}{b}\)
=> \(\frac{b}{c}=\frac{a}{b}\Rightarrow b^2=ac\)
khi đó: \(\frac{a^2+b^2}{b^2+c^2}=\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)
Vậy:...
Bài 2 sau khi đã sửa đề thành $5x=7z$:
Ta có:
\(\frac{x}{y}=\frac{3}{2}\Leftrightarrow \frac{x}{3}=\frac{y}{2}\Leftrightarrow \frac{x}{21}=\frac{y}{14}(1)\)
\(5x=7z\Leftrightarrow \frac{x}{7}=\frac{z}{5}\Leftrightarrow \frac{x}{21}=\frac{z}{15}(2)\)
Từ $(1);(2)\Rightarrow \frac{x}{21}=\frac{y}{14}=\frac{z}{15}$ và đặt bằng $k$
$\Rightarrow x=21k; y=14k; z=15k$
Khi đó:
$x-2y+z=32$
$\Leftrightarrow 21k-28k+15k=32\Leftrightarrow 8k=32\Rightarrow k=4$
$\Rightarrow x=21k=84; y=14k=56; z=15k=60$
Bài 2: $5z=7z$ hình như sai, bạn coi lại đề.
Bài 3:
\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Leftrightarrow \frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)
\(\Leftrightarrow \frac{9a+(a+b)}{a+b}=\frac{9b+(b+c)}{b+c}\Leftrightarrow \frac{9a}{a+b}+1=\frac{9b}{b+c}+1\)
\(\Leftrightarrow \frac{a}{a+b}=\frac{b}{b+c}\Rightarrow ab+ac=ab+b^2\)
\(\Leftrightarrow ac=b^2\Rightarrow \frac{a}{b}=\frac{b}{c}\) (đpcm)
Ta có : \(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\)
\(\Rightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)
\(\Rightarrow\frac{9a}{a+b}=\frac{9b}{b+c}\Rightarrow\frac{a}{a+b}=\frac{b}{b+c}\)
=> a(b + c) = b(a + b)
=> ab + ac = ab + bb
=> ac = bb
=> \(\frac{a}{b}=\frac{b}{c}\left(\text{đpcm}\right)\)