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Ta có \(\frac{n+2}{n-2}=\frac{m+3}{m-3}\Leftrightarrow1+\frac{4}{n-2}=1+\frac{6}{n-2}\)
\(\Leftrightarrow\frac{4}{n-2}=\frac{6}{m-3}\Leftrightarrow4\left(m-3\right)=6\left(n-2\right)\)
\(\Leftrightarrow4m-12=6n-12\)
\(\Leftrightarrow4m=6n\Leftrightarrow2m=3n\)
\(\Leftrightarrow\frac{n}{2}=\frac{m}{3}\left(đpcm\right)\)
Hok tốt
bạn xem lại đề:
Có \(\frac{3}{2}\frac{3+7}{2+7}=\frac{10}{9}\)
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(=>\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(=>\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(=>m=5\)
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(=>\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(=>\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(=>n=3\)
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m =5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
Ta có: \(m+n\ne0.\)
\(\Rightarrow m+n+2017\ne2017.\)
Có:
\(x=\frac{m}{n+2017}=\frac{n}{m+2017}=\frac{2017}{m+n}\) và \(m+n+2017\ne2017.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(x=\frac{m}{n+2017}=\frac{n}{m+2017}=\frac{2017}{m+n}\)
\(\Rightarrow x=\frac{m+n+2017}{n+2017+m+2017+m+n}\)
\(\Rightarrow x=\frac{m+n+2017}{2m+2n+4034}\)
\(\Rightarrow x=\frac{m+n+2017}{2.\left(m+n+2017\right)}\)
\(\Rightarrow x=\frac{1}{2}.\)
Vậy \(x=\frac{1}{2}.\)
Chúc bạn học tốt!
Các bạn giúp ạ : @Vũ Minh Tuấn , @Băng Băng 2k6 , @Phạm Lan Hương , và cô @Akai Haruma
a) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
=> m = 5
Vậy m = 5
b) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Rightarrow\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
=> n = 3
Vậy n = 3
đặt: m/n=p/q=k
suy ra: m=kn; p=kq
Suy ra: \(\hept{\begin{cases}VT=\frac{n}{3n+kn}=\frac{1}{3+k}\\VP=\frac{q}{3q+kq}=\frac{1}{3+k}\end{cases}\Rightarrow VT=VP\left(ĐPCM\right)}\)
Sửa đề:CM:\(\left(p-m\right)^2=4\left(m-n\right)\left(n-p\right)\)
Ta có:\(\frac{m}{2014}=\frac{n}{2015}=\frac{p}{2016}=\frac{p-m}{2016-2014}=\frac{p-m}{2}=\frac{m-n}{2014-2015}\)=
\(=\frac{m-n}{-1}=\frac{n-p}{2014-2016}=\frac{n-p}{-1}\)
\(\Rightarrow\frac{\left(p-m\right)^2}{4}=\frac{\left(m-n\right).\left(n-p\right)}{\left(-1\right).\left(-1\right)}\)
\(\Rightarrow\frac{\left(p-m\right)^2}{4}=\frac{\left(m-n\right)\left(n-p\right)}{1}\)
\(\Rightarrow\left(p-m\right)^2=4\left(m-n\right)\left(n-p\right)\)
\(\frac{m}{n}< \frac{p}{q}\Leftrightarrow mq< pn\)
\(\Rightarrow mq+mn< pn+mn\)
\(\Rightarrow m\left(n+q\right)< n\left(m+p\right)\)
\(\Rightarrow\frac{m}{n}< \frac{m+p}{n+q}\) (1)
\(mq< pn\)
\(\Rightarrow mq+pq=pn+pq\)
\(\Rightarrow q\left(m+p\right)< p\left(n+q\right)\)
\(\Rightarrow\frac{m+p}{n+q}< \frac{p}{q}\) (2)
Từ (1) và (2) suy ra: \(\frac{m}{n}< \frac{m+p}{n+q}< \frac{p}{q}\)