Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1^3}{2^3}\right)^5\)
\(\left(\frac{1}{2}\right)^n=\left[\left(\frac{1}{2}\right)^3\right]^5\)
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)
n = 15
\(\left(\frac{1}{2}\right)^n=\left(\frac{1}{8}\right)^5\)
\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{3.5}\)
\(\Rightarrow\left(\frac{1}{2}\right)^n=\left(\frac{1}{2}\right)^{15}\)
\(\Rightarrow n=15\)
Vậy n = 15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{5^{32}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Có \(P\left(1\right)=2.1^2+2m.1+m^2=2+2m+m^2\)
\(Q\left(1\right)=\left(-1\right)^2+4\left(-1\right)+5=1-4+5=2\). Vì \(P\left(1\right)=Q\left(-1\right)\)
\(\Rightarrow2+2m+m^2=2\Leftrightarrow2m+m^2=2-2=0\Leftrightarrow m\left(2+m\right)=0\)
\(\Rightarrow m=0\) hoặc \(2+m=0\Leftrightarrow m=0-2=-2\)
b) Đặt \(Q\left(x\right)=x^2+4x+5=0\Leftrightarrow x^2+4x=0-5=-5\)
\(\Leftrightarrow x\left(x+4\right)=-5\). Từ đó bạn lập bảng ra sẽ thấy k có trường hợp thỏa mãn => Vô nghiệm