Tìm số dư 109^345 / 7
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Ta có : \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
\(\Rightarrow\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
=> \(\frac{a}{c}=\frac{b}{d}\)
=> \(\frac{a}{b}=\frac{c}{d}\) nếu khố hiểu thì bạn chứng mình kiểu này :
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
=> \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
Mặt khác \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
=> \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)
Vậy \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2017-Ix-2017I=x
\(TH1:x>0\)
\(\Rightarrow2017-\left(x-2017\right)=x\)
\(\Rightarrow2017-x+2017\)
\(\Rightarrow4034-x=x\) ( loại )
\(TH2:x\le0\)
\(\Rightarrow2017-\left[-\left(x-2017\right)\right]=x\)
\(\Rightarrow2017-\left(-x+2017\right)=x\)
\(\Rightarrow2017+x-2017=x\)
\(\Rightarrow x+0=x\)
\(\Rightarrow x=x\) \(\left(x\in Z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+\frac{2}{3}\right)^3=\frac{1}{8}\)
\(\Rightarrow x+\frac{2}{3}=\frac{1}{2}\)
\(x=\frac{1}{2}-\frac{2}{3}\)
\(x=\frac{-1}{6}\)
b) 52x-1-125 = 0
52x-1 = 0+125
52x-1 = 125
<=> 52x-1 = 53
=> 2x-1=3
=> x = 2
c) \(\frac{8^1}{3^{2x+1}}=3\)
\(\Rightarrow8=3.3^{2x+1}=3^{2x+1+1}=3^{2x+2}\)
\(\Rightarrow8\ne3^{2x+2}\)
=> x vô nghiệm
a, \(\left(\frac{x+2}{3}\right)^3=\frac{1}{8}\)\(\Rightarrow\left(\frac{x+2}{3}\right)^3=\left(\frac{1}{2}\right)^3\)
\(\Rightarrow\frac{x+2}{3}=\frac{1}{2}\)\(\Rightarrow\left(x+2\right).2=3.1\)\(\Rightarrow x+2=\frac{3}{2}\)\(\Rightarrow x=-\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a_1-1}{9}=\frac{a_2-2}{8}=...=\frac{a_9-9}{1}\)
\(=\frac{a_1-1+a_2-2+...+a_9-9}{9+8+...+1}\)\(=\frac{\left(a_1+a_2+...+a_9\right)-\left(1+2+...+9\right)}{45}\)\(=\frac{90-45}{45}=1\)
Do dó, suy ra:\(\frac{a_1-1}{9}=1\Rightarrow a_1=10\)
\(\frac{a_2-2}{8}=1\Rightarrow a_2=10\)
\(...\)
\(\frac{a_9-9}{1}=1\Rightarrow a_9=10\)
Vậy \(a_1=a_2=...=a_9=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo tính chất tỉ dãy số bằng nhau thì:
\(\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}=1\)
\(\Leftrightarrow\frac{a+b}{c+d}=\frac{b+c}{d+a}=\frac{c+d}{a+b}=\frac{d+a}{b+c}=1\)
\(\Rightarrow M\Leftrightarrow1+1+1+1=4\)
Ps: Cách mình nhanh hơn nè!