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\(\frac{3a+7b}{7a+3b}=1\)
<=> 3a + 7b = 7a + 3b
<=> 3a - 7a = 3b - 7b
<=> -4a = -4b
<=> a = b
Thay a = b vào, ta có:
A = - a + a - 1
=> A = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3a+7b}{7a+3b}=1\)
\(\Rightarrow\)3a+7b=7a+3b
\(\Rightarrow\)7b-3b=7a-3a(chuyển vế đổi dấu)
\(\Rightarrow\)4b=4a
\(\Rightarrow\)b=a
\(\Rightarrow\)b-a=0
hay -a+b=0
\(\Rightarrow\)-a+b-1=0-1= -1
Hay A= -1
Tik mik nha!
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a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
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Đặt \(\frac{a}{7}=\frac{b}{3}=k\left(k\ne0\right)\)
\(\Rightarrow a=7k\), \(b=3k\)
Thay a, b vào biểu thức A ta được
\(A=\frac{4.7k-7.3k}{3.7k-5.3k}=\frac{28k-21k}{21k-15k}=\frac{7k}{6k}=\frac{7}{6}\)
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\(a-b=13\Rightarrow a=b+13\)
thay \(a=b+13\) vào biểu thức thì ta có:
\(\frac{3a-b}{2a+13}-\frac{3b-a}{2b-13}=\frac{3\left(b+13\right)-b}{2\left(b+13\right)+13}-\frac{3b-\left(b+13\right)}{2b-13}\)
\(=\frac{2b+39}{2b+39}-\frac{2b-13}{2b-13}=1-1=0\)
\(\frac{3a+7b}{7a+3b}=1\)
=> 3a + 7b = 7a + 3b
=> 3a - 3b = 7a - 7b
=> 3(a - b) = 7(a - b)
=> 7(a - b) - 3(a - b) = 0
=> 4(a - b) = 0
=> a - b = 0
=> a = b
Ta có : A = -a + b - 1
=> A = -a + a - 1
=> A = -1
cick mình nha bạn
3a+7b=7a+3b => a=b
-a+b-1=-a+a-1=-1