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soyeon_Tiểubàng giải1 tháng 10 2016 lúc 20:35
Ta có:
5a + 7b/6a + 5b = 29/28
=> (5a + 7b).28 = (6a + 5b).29
=> 140a + 196b = 174a + 145b
=> 196b - 145b = 174a - 140a
=> 51b = 34a
=> 3b = 2a
=> a/b = 3/2
Mà (a,b)=1; a,b thuộc N
=> a = 3; b = 2
Vậy a = 3; b = 2
bài của bạn Nguyễn Văn Hòa hợp con nhà bà lý luôn
=) =) =) =) =)
\(\frac{5a+7b}{6a+5b}=\frac{29}{28}\)
=>(5a+7b)28=(6a+5b)29
=>140a+196b=174a+145b
=>(196 - 145)b=(174 - 140 )a
=>51b=34a
=>\(\frac{a}{b}=\frac{51}{34}=\frac{3}{2}\)
=>a=3k
b=2k
\(\left(k\in N\right)\)
Mà (2;3)=1
(a;b)=1
=>K=1
=>a=3
b=2
\(28\left(5a+7b\right)=29\left(6a+5b\right)\Leftrightarrow140a+196b=174a+145b\Leftrightarrow51b=34a\Leftrightarrow\frac{a}{b}=\frac{51}{34}=\frac{3}{2}\)
Vì (a;b) =1
=> a =3 ; b =2
\(\frac{5a+7b}{6a+5b}=\frac{29}{28}\Rightarrow28\left(5a+7b\right)=29\left(6a+5b\right)\Rightarrow140a+196b=174a+145b\)
\(\Rightarrow\left(140a+196b\right)-\left(174a+145b\right)=0\Rightarrow140a+196b-174a-145b=0\)
\(\Rightarrow140a-174a+196b-145b=0\Rightarrow\left(-34\right)a+51b=0\)
\(\Rightarrow51b=34a\Rightarrow\frac{a}{51}=\frac{b}{34}\)
G/S:\(\frac{a}{51}=\frac{b}{34}=k\Rightarrow a=51k;b=34k\)
\(\frac{5a+7b}{6a+5b}=\frac{29}{28}\Rightarrow\frac{5.51k+7.34k}{6.51k+5.34k}=\frac{255k+238k}{306k+170k}=\frac{493k}{476k}=\frac{29k}{28k}=\frac{29}{28}\Rightarrow k=1\)
\(\Rightarrow a=51.1=51;b=34.1=34\)
D/S: ........
\(\Leftrightarrow28\left(5a+7b\right)=29\left(6a+5b\right)\Leftrightarrow140a+196b=174a+145b\Leftrightarrow51b=34a\Leftrightarrow\frac{a}{b}=\frac{51}{34}=\frac{3}{2}\)
Vì (a;b) =1
=> a =3 ; b =2
\(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)
\(\Rightarrow28\left(5a+7b\right)=29\left(6a+5b\right)\)
\(\Rightarrow140a+196b=174a+145b\)
\(\Rightarrow140a=174a+145b-196b\)
\(\Rightarrow140a=174a-51b\)
\(\Rightarrow34a=51b\)
\(\Rightarrow2a=3b\)
Mà \(\left(a,b\right)=1\) nên \(a=3;b=2\)
\(\dfrac{5a+7b}{6a+5b}=\dfrac{29}{28}\)
\(\Rightarrow140a+196b=174a+145b\)
\(\Rightarrow51b=34a\)
\(\Rightarrow3b=2a\)
Do \(\left(a;b\right)=1\)
\(\Rightarrow a=3,b=2\)
Vậy...
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
Theo đề: \(\frac{a}{5b}=\frac{b}{5c}=\frac{c}{5d}=\frac{d}{5a}\)
=> \(\frac{1}{5}.\frac{a}{b}=\frac{1}{5}.\frac{b}{c}=\frac{1}{5}.\frac{c}{d}=\frac{1}{5}.\frac{d}{a}\)
=> \(\frac{1}{5}.\frac{a}{b}.5=\frac{1}{5}.\frac{b}{c}.5=\frac{1}{5}.\frac{c}{d}.5=\frac{1}{5}.\frac{d}{a}.5\)
=> \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=\frac{a+b+c+d}{b+c+d+a}=1\)(do \(a+b+c+d\ne0\))
Từ \(\frac{a}{b}=1\Rightarrow a=b\)(1)
Từ \(\frac{b}{c}=1\Rightarrow b=c\)(2)
Từ \(\frac{c}{d}=1\Rightarrow c=d\)(3)
Từ \(\frac{d}{a}=1\Rightarrow d=a\)(4)
Từ (1), (2), (3) và (4) suy ra : a = b = c = d (đpcm)
Giả sử \(a>b\),ta có:
\(\frac{a}{5b}=\frac{b}{5c}\Rightarrow5b>5c\Rightarrow b>c\)vì \(a>b\)
\(\frac{b}{5c}=\frac{c}{5d}\Rightarrow5c>5d\Rightarrow c>d\)vì \(b>c\)
\(\frac{c}{5d}=\frac{d}{5a}\Rightarrow5d>5a\Rightarrow d>a\)vì \(c>d\)
Từ 4 dòng trên \(\Rightarrow a>b>c>d\)
\(\frac{a}{5b}=\frac{d}{5a}\Rightarrow5b< 5a\Rightarrow b< a\)vì \(a>d\)
\(\Rightarrow\)Với \(a>b\)thì không thỏa mãn.
Chứng minh tương tự với \(a< b\)thì ta lại thấy vô lý vì \(a>b\)
\(a>b;a< b\)vô lý thì \(a=b\)thỏa mãn.
\(\frac{a}{5b}=\frac{b}{5c}\Rightarrow5b=5c\Rightarrow b=c\)vì \(a=b\)
\(\frac{b}{5c}=\frac{c}{5d}\Rightarrow5c=5d\Rightarrow c=d\)vì \(b=c\)
\(\frac{c}{5d}=\frac{d}{5a}\Rightarrow5d=5a\Rightarrow d=a\)vì \(c=d\)
Theo tính chất Bắc-Cầu thì ta kết luận được \(a=b=c=d\left(đpcm\right)\)