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\(\frac{a+1}{2}\)=\(\frac{b+2}{3}\)=\(\frac{c+2}{4}\)=>\(\frac{3a+3}{6}\)=\(\frac{-2b-4}{-6}\)=\(\frac{c+2}{4}\)=\(\frac{3a-2b+c+3-4+2}{6+-6+4}\)=\(\frac{105+3-4+2}{4}\)=26,5
=>a=26,5.2-1=52
b=26,5.3-2=77,5
c=26,5.4-2=104
chắc chắn đúng
-duc tuan nguyen-
=>( 3a + 3 ) / 6 = ( 2b + 4 ) / 6 =( c + 2 ) / 4
= ( 3a + 3 - 2b - 4 + c + 2 ) / ( 6 - 6 + 4 )
= 106 / 4
= 26,5
=> a + 1 = 53 => a = 52
=> b + 2 = 79,5 => b = 77,5
=> c + 2 = 106 => c = 104
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\(\frac{1}{a+2}=\frac{2}{a+6}\)
\(\Rightarrow x+6=2\left(a+2\right)\)
\(\Rightarrow x+6=2x+4\)
\(\Rightarrow-x=-2\)
\(\Rightarrow x=2\)
a) \(\frac{1}{a+2}=\frac{2}{a+6}\)
=> a + 6 = 2(a + 2)
=> a + 6 = 2a + 4
=> a - 2a = 4 - 6
=> -a = -2
=> a = 2
c) \(\frac{3a-7}{a-1}=2\)
=> 3a - 7 = 2(a - 1)
=> 3a - 7 = 2a - 2
=> 3a - 2a = -2 + 7
=> a = 5
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Ta có :
\(\frac{2-2a}{6-8b}=\frac{2\left(1-a\right)}{2\left(3-4b\right)}=\frac{1-a}{3-4b}\)
\(\frac{3-3a}{9-12b}=\frac{3\left(1-a\right)}{3\left(3-4b\right)}=\frac{1-a}{3-4b}\)
\(\Rightarrow\frac{2-2a}{6-8b}=\frac{3-3a}{9-12b}\) (đpcm)
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Ta có : a + b = 7 \(\Rightarrow\)a = 7 - b
b = 7 - a
Thế vào câu trên ta có :
\(\frac{3a+b}{2a+7}=\frac{3\left(7-b\right)+b}{2\left(7-b\right)+7}=\frac{21-3b+b}{14-2b+7}=\frac{21-2b}{21-2b}=1\)
\(\frac{2a+b}{a+7}=\frac{2\left(7-b\right)+b}{7-b+7}=\frac{14-2b+b}{14-b}=\frac{14-b}{14-b}=1\)
Vậy \(\frac{3a+b}{2a+7}-\frac{2a+b}{a+7}=1-1=0\)
\(\frac{3a+b}{2a+7}=\frac{3a+b}{2a+a+b}=\frac{3a+b}{3a+b}=1\)(1)
\(\frac{2a+b}{a+7}=\frac{2a+b}{a+a+b}=\frac{2a+b}{2a+b}=1\)(2)
Từ (1) và (2) => ĐPCM