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\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab-5b+6a-30\)
\(\Leftrightarrow-6a+5b=6a-5b\)
\(\Leftrightarrow5b+5b=6a+6a\)
\(\Leftrightarrow10b=12a\)
\(\Leftrightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)
a ,A = \(a.\frac{1}{3}+a.\frac{1}{4}-a.\frac{1}{6}\)
\(=a.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\\ =\frac{-3}{5}.\frac{5}{12}\)
\(=\frac{-1}{4}\)
b, B = \(b.\frac{5}{6}+b.\frac{3}{4}-b.\frac{1}{2}\)
\(=b.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\frac{7}{12}\)
\(=\frac{7}{13}\)
a) Thay \(a=\frac{-3}{5}\)vào biểu thức A ta có :
\(A=\frac{-3}{5}.\frac{1}{3}+\frac{-3}{5}.\frac{1}{4}-\frac{-3}{5}.\frac{1}{6}\)
\(A=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(A=\frac{-3}{5}.\frac{5}{12}\)
\(A=\frac{-1}{4}\)
Vậy giá trị của biểu thức A tại \(a=\frac{-3}{5}\)là \(\frac{-1}{4}\)
b) Thay \(b=\frac{12}{13}\)vào biểu thức B ta có :
\(B=\frac{12}{13}.\frac{5}{6}+\frac{12}{13}.\frac{3}{4}-\frac{12}{13}.\frac{1}{2}\)
\(B=\frac{12}{13}.\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(B=\frac{12}{13}.\frac{13}{12}\)
\(B=1\)
Vậy giá trị của biểu thức B tại \(b=\frac{12}{13}\)là 1
_Chúc bạn học tốt_
Đặt \(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}=k\Rightarrow\hept{\begin{cases}a=5k\\b=6k\\c=7k\end{cases}}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(5k-6k\right)\left(6k-7k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)(1)
và \(\left(c-a\right)^2=\left(7k-5k\right)^2=\left(2k\right)^2=4k^2\)(2)
Từ (1) và (2) suy ra \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có;
\(\frac{a}{5}=\frac{b}{6}=\frac{c}{7}=\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Leftrightarrow\hept{\begin{cases}a-b=b-c\\c-a=-2\left(b-c\right)=-2\left(a-b\right)\end{cases}}\)
\(\left(c-a\right)^2=-2\left(a-b\right)\cdot-2\left(b-c\right)=4\left(a-b\right)\left(b-c\right)\)(đpcm)
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
d) \(x.\left(y+2\right)-y=15\)
\(\Rightarrow x.\left(y+2\right)=15+y\)
\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)
y + 2 là ước nguyên của 13
\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)
\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)
\(y+2=13\Rightarrow y=11\Rightarrow x=2\)
\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)
Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu
\(A=5\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\right)\)
\(=5\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\right)\)
\(=5\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{81-80}{80.81}\right)\)
\(=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\right)\)
\(=5\left(1-\frac{1}{81}\right)=\frac{5.80}{81}=\frac{400}{81}\)
b)
\(B=7\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{483}\right)\)
\(=7.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{21.23}\right)\)
=> \(2.B=7\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{21.23}\right)\)
\(=7\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{23-21}{21.23}\right)\)
\(=7.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{21}-\frac{1}{23}\right)\)
\(=7\left(\frac{1}{3}-\frac{1}{23}\right)=\frac{7.20}{69}=\frac{140}{69}\)
=> \(B=\frac{140}{69}:2=\frac{70}{69}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}=>\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(=>a\left(b-6\right)+5\left(b-6\right)=a\left(b+6\right)-5\left(b+6\right)\)
\(=>ab-6a+5b-30=ab+6a-5b-30=>-6a+5b=6a-5b=>6a-\left(-6a\right)=5b-\left(-5b\right)\)
\(=>12a=10b=>\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\) (đpcm)