Tìm ba số a,b,c biết 5a-4b/6=6b-5c/4=4c-6a/5 và a+b+c=45
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{5a-4b}{6}=\frac{6b-5c}{4}=\frac{4c-6a}{5}\)
\(\Rightarrow \frac{6(5a-4b)}{36}=\frac{4(6b-5c)}{16}=\frac{5(4c-6a)}{25}=\frac{6(5a-4b)+4(6b-5c)+5(4c-6a)}{36+16+25}=\frac{0}{36+16+25}=0\)
\(\Rightarrow \left\{\begin{matrix} 5a-4b=0\\ 6b-5c=0\\ 4c-6a=0\end{matrix}\right.\Rightarrow \frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)
Áp dụng TCDTSBN:
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{6}=\frac{a+b+c}{4+5+6}=\frac{45}{15}=3\)
\(\Rightarrow \left\{\begin{matrix} a=4.3=12\\ b=5.3=15\\ c=5.3=18\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\left(5a-4b\right)6}{36}=\frac{\left(6a-4c\right)5}{25}=\frac{\left(6b-5c\right)4}{16}=\frac{\left(5a-4b\right)6-\left(6a-4c\right)5+\left(6b-5c\right)4}{36-25+16}=\frac{0}{27}\)
\(\Rightarrow5a=4b\Leftrightarrow\frac{a}{4}=\frac{b}{5}\)
\(\Rightarrow6a=4c\Leftrightarrow\frac{a}{4}=\frac{c}{6}\)
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=(-5c+3a-4b)-(3a-4b+7c)-(-12b-6a+15c)+(-3c+21a-10b)
=-5c+3a-4b-3a+4b-7c+12b+6a-15c
=6a +12b -27c
C=-(-32b-12c+5a)+(2c-4b-23a)-(17a-16c-31b)-(-6b+3c)
=32b+12c-5a+2c-4b-23a-17a+16c+31b+6b-3c
=-45a+65b+9c
![](https://rs.olm.vn/images/avt/0.png?1311)
phá ngoặc lun nà
+4a-5c+3b-2b+a-7c-7b+3c-5a=(4a+a-5a)+(3b-2b-7b)+(-5c-7c+3c)=0-6b-9c=-9c-6b
-2a+3c-b-5b-4c+12a+9b+4c-4a-6a-3b-3c+d=(-2a+12a-4a-6a)+(-b-5b+9b-3b)+(3c-4c+4c-3c)+d=0+0+0+0+d=d
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}2a+3b+2c=5\\5a+4b+c=55\\a+b-4c=24\end{cases}}\Leftrightarrow8a+8b-c=5+55+24\)
\(\Leftrightarrow8a+8b-c=84\)
\(\Leftrightarrow8\left(a+b\right)-c=84\)
\(\Leftrightarrow8\left(a+b\right)=84+c\)
\(\Rightarrow a+b+c=84\)
\(\Rightarrow TBC\left(a,b,c\right)=\frac{84}{3}=28\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2 : a) \(\dfrac{a-1}{2}=\dfrac{b+3}{4}=\dfrac{c-5}{6}\)
áp dụng dảy tỉ số bằng nhau
ta có : \(\dfrac{5\left(a-1\right)-3\left(b+3\right)-4\left(c-5\right)}{5.2-3.4-4.6}\)
\(=\dfrac{5a-5-3b-9-4c+20}{10-12-24}=\dfrac{\left(5a-3b-4c\right)-5-9+20}{-26}\)
\(=\dfrac{46+6}{-26}=\dfrac{52}{-26}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a-1}{2}=-2\\\dfrac{b+3}{4}=-2\\\dfrac{c-5}{6}=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-1=-4\\b+3=-8\\c-5=-12\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-11\\c=-7\end{matrix}\right.\)
vậy \(a=-3;b=-11;c=-7\)
b) ta có : \(3a=2b\Leftrightarrow6a=4b=5c\Leftrightarrow\dfrac{6a}{2}=\dfrac{4b}{2}=\dfrac{5c}{2}\)
áp dụng dảy tỉ số bằng nhau
ta có \(\dfrac{-60a-60b+60c}{-10.2-15.2+12.2}=\dfrac{60\left(-a-b+c\right)}{-20-30+24}\)
\(=\dfrac{60\left(-52\right)}{-26}=\dfrac{-3120}{-26}=120\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{6a}{2}=120\\\dfrac{4b}{2}=120\\\dfrac{5c}{2}=120\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6a=240\\4b=240\\5c=240\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\\c=48\end{matrix}\right.\)
vậy \(a=40;b=60;c=48\)