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Ta có : 4a = 3b => 28a = 21b (1)
7b = 5c => 21b = 15c (2)
Từ (1) và (2) => 28a = 21b = 15c
Ta có : 28a = 21b = 15c \(=\frac{a}{\frac{1}{28}}=\frac{b}{\frac{1}{21}}=\frac{c}{\frac{1}{15}}=\frac{2a}{\frac{1}{14}}=\frac{3b}{\frac{1}{7}}=\frac{2a+3b-c}{\frac{1}{14}+\frac{1}{7}-\frac{1}{15}}=\frac{186}{\frac{31}{210}}=1260\)
Nên : 28a = 1260 => a = 45
21b = 1260 => b = 60
15c = 1260 => c = 84
Vậy ........................
Ta có:
\(4a=3b\)=> \(\frac{a}{3}=\frac{b}{4}\)=> \(\frac{a}{15}=\frac{b}{20}\left(1\right)\)
\(7b=5c\)=>\(\frac{b}{5}=\frac{c}{7}\) => \(\frac{b}{20}=\frac{c}{28}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\)
=>\(\frac{a}{15}=\frac{b}{20}=\frac{c}{28}\)=>\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2a}{30}=\frac{3b}{60}=\frac{c}{28}=\frac{2a+3b-c}{30+60-28}=\frac{186}{62}=3\)
=>\(\frac{a}{15}=3\)=>\(a=45\)
\(\frac{b}{20}=3\)=>\(b=60\)
\(\frac{c}{28}=3\)=>\(c=84\)
Vậy \(a=40;b=60;c=84\)
Ta có: \(2a=3b\)=> \(\frac{a}{3}=\frac{b}{2}\)=>\(\frac{a}{21}=\frac{b}{14}\left(1\right)\)
\(5b=7c\)=>\(\frac{b}{7}=\frac{c}{5}\) =>\(\frac{b}{14}=\frac{c}{10}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\)
=>\(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)=> \(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{3a}{63}=\frac{7b}{98}=\frac{5c}{50}=\frac{3a-7b+5c}{63-98+50}=\frac{30}{15}=2\)
=>\(\frac{a}{21}=2\)=>\(a=42\)
\(\frac{b}{14}=2\)=>\(b=28\)
\(\frac{c}{10}=2\)=>\(c=20\)
Vậy \(a=42;b=28;c=20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(2a=5b;7b=9c\)
\(2a=5b\Rightarrow\dfrac{a}{5}=\dfrac{b}{2}\Rightarrow\dfrac{a}{45}=\dfrac{b}{18}\)
\(7b=9c\Rightarrow\dfrac{b}{9}=\dfrac{c}{7}\Rightarrow\dfrac{b}{18}=\dfrac{c}{14}\)
\(\Rightarrow\dfrac{a}{45}=\dfrac{b}{18}=\dfrac{c}{14}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{45}=\dfrac{b}{18}=\dfrac{c}{14}=\dfrac{4a}{4.45}=\dfrac{5b}{5.18}=\dfrac{8c}{8.14}=\dfrac{4a-5b+8c}{180-90+112}=\dfrac{404}{202}=2\)
\(\dfrac{a}{45}=2\Rightarrow a=45.2=90\)
\(\dfrac{b}{18}=2\Rightarrow b=18.2=36\)
\(\dfrac{c}{14}=2\Rightarrow c=14.2=28\)
Vậy \(a=90\) và \(b=36\) và \(c=28\)
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ta co 4a = 5b
\(\Rightarrow\frac{a}{5}=\frac{b}{4}\)
\(\Rightarrow\frac{2a}{10}=\frac{b}{4}\)
áp dung tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{2a}{10}=\frac{b}{4}=\frac{2a-b}{10-4}=\frac{9}{6}=\frac{3}{2}\)
\(\Rightarrow\frac{2a}{10}=\frac{3}{2}\Rightarrow a=7,5\)
\(\frac{b}{4}=\frac{3}{2}\Rightarrow b=6\)
vậy a=7,5 ; b=6
Có b -2a = 9
b = 9 + 2a
Lại có 4a = 5b
4a - 5b = 0
4a - 45 - 10a =0
-6a = 45
a = -7,5
khi đó b = 9 - 2 . 7,5 = -6
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\dfrac{a}{b}=\dfrac{3}{4}\Leftrightarrow\dfrac{a}{3}=\dfrac{b}{4}\\ Đặt\dfrac{a}{3}=\dfrac{b}{4}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\end{matrix}\right.\\ ThayvàoA,tacó:\)
\(A=\dfrac{2a-5b}{a-3b}-\dfrac{4a+b}{8a-2b}\\ \Leftrightarrow=\dfrac{2\cdot3k-5\cdot4k}{3k-3\cdot4k}-\dfrac{4\cdot3k+4k}{8\cdot3k-2\cdot4k}\\ =\dfrac{6k-20k}{3k-12k}-\dfrac{12k+4k}{24k-8k}\\ =\dfrac{14k}{9k}-\dfrac{16k}{16k}\\ =\dfrac{14}{9}-1\\ =\dfrac{5}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lớp 7 gì mà dễ ẹc :))
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Leftrightarrow6a-3b=2a+2b\)
\(\Rightarrow4a=5b\)
\(\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Leftrightarrow4a-2b=3b-3c+3a\)
\(\Leftrightarrow a=5b-3c\)
\(\Leftrightarrow a-5b=-3c\)
\(\Leftrightarrow a-4a=-3c\)
\(\Leftrightarrow-3a=-3c\)
\(\Rightarrow a=c\)
Ta có : \(P=\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(4a+4a\right)^5}{\left(4a+4a\right)^2\left(a+3a\right)^3}=\frac{\left(8a\right)^3}{\left(4a\right)^3}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{2}{3}\)
\(\Rightarrow\frac{2a-b}{a+b}=\frac{b-c+a}{2a-b}=\frac{\left(2a-b\right)+\left(b-c+a\right)}{\left(a+b\right)+\left(2a-b\right)}=\frac{3a-c}{3a}=\frac{2}{3}\)
\(\Rightarrow2\times3a=3\times\left(3a-c\right)\)
\(\Rightarrow6a=9a-3c\)
\(\Rightarrow6a-9a=-3c\)
\(\Rightarrow-3a=-3c\)
\(\Rightarrow\frac{-3a}{-3}=\frac{-3c}{-3}\)
\(\Rightarrow a=c\)
\(\Rightarrow\frac{\left(5b+4a\right)^5}{\left(5b+4c\right)^2\left(a+3c\right)^3}=\frac{\left(5b+4a\right)^5}{\left(5b+4a\right)^2\left(a+3a\right)^3}=\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}\)
\(\frac{2a-b}{a+b}=\frac{2}{3}\)
\(\Rightarrow3\times\left(2a-b\right)=2\left(a+b\right)\)
\(\Rightarrow6a-3b=2a+2b\)
\(\Rightarrow6a-2a=3b+2b\)
\(\Rightarrow4a=5b\)
\(\Rightarrow b=\frac{4a}{5}\)
\(\Rightarrow\frac{\left(5b+4a\right)^3}{\left(4a\right)^3}=\left(\frac{5\times\frac{4a}{5}+4a}{4a}\right)^3=\left(\frac{4a+4a}{4a}\right)^3\)
\(\Rightarrow\left(\frac{8a}{4a}\right)^3=2^3=8\)
4a = 5b
\(\Rightarrow\dfrac{a}{5}=\dfrac{b}{4}\)
\(\Rightarrow\dfrac{2a}{10}=\dfrac{b}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\dfrac{2a}{10}=\dfrac{b}{4}=\dfrac{b-2a}{4-10}=\dfrac{-5}{-6}=\dfrac{5}{6}\)
\(\Rightarrow\left\{{}\begin{matrix}2a=\dfrac{5}{6}.10\\b=\dfrac{5}{6}.4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{25}{6}\\b=\dfrac{10}{3}\end{matrix}\right.\)