Không dùng máy tính cầm tay, hãy tính \({\log _9}\frac{1}{{27}}.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
B
1
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(log_50,5=-0,439677\)
c) \(In\left(\dfrac{3}{2}\right)=0,405465\)
DH
2
H
30 tháng 4 2022
Tham khảo
A=3/2+7/6+13/12+...+91/90
A=1+1/2+1+1/6+…+1+1/72+1+1/90
A=(1+1+1+…+1+1)+1/1.2+1/2.3+1/3.4+…+1/9.10
A=10+1/1-1/2+1/2-1/3+…-1/9+1/9+1/10
A=10+1-1/10
A=10+9/10
A=109/10
30 tháng 4 2022
\(S=\dfrac{3}{2}+\dfrac{7}{6}+\dfrac{13}{12}+...+\dfrac{91}{90}\)
\(=1+\dfrac{1}{2}+1+\dfrac{1}{6}+1+\dfrac{1}{12}+...+1+\dfrac{1}{90}\)
\(=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)+9\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)+9\)
\(=1-\dfrac{1}{10}+9=\dfrac{99}{10}\)
HQ
Hà Quang Minh
Giáo viên
25 tháng 8 2023
\(a,cos\left(\dfrac{5\pi}{12}\right)=cos\left(\dfrac{\pi}{4}+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{4}\right)cos\left(\dfrac{\pi}{6}\right)-sin\left(\dfrac{\pi}{4}\right)sin\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\ sin\left(\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{4}+\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}\right)cos\left(\dfrac{\pi}{6}\right)+cos\left(\dfrac{\pi}{4}\right)sin\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ tan\left(\dfrac{5\pi}{12}\right)=\dfrac{sin\left(\dfrac{5\pi}{12}\right)}{cos\left(\dfrac{5\pi}{12}\right)} =2-\sqrt{3}\\ cot\left(\dfrac{5\pi}{12}\right)=\dfrac{1}{tan\left(\dfrac{5\pi}{12}\right)}=\dfrac{1}{2-\sqrt{3}}\)
\(b,cos\left(-555^o\right)=cos\left(3\pi+\dfrac{\pi}{12}\right)=-cos\left(\dfrac{\pi}{12}\right)=-cos\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=-\left[cos\left(\dfrac{\pi}{3}\right)cos\left(\dfrac{\pi}{4}\right)+sin\left(\dfrac{\pi}{3}\right)sin\left(\dfrac{\pi}{4}\right)\right]=-\dfrac{\sqrt{6}+\sqrt{2}}{4}\\ sin\left(-555^o\right)=sin\left(3\pi+\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=sin\left(\dfrac{\pi}{3}\right)cos\left(\dfrac{\pi}{4}\right)-cos\left(\dfrac{\pi}{3}\right)sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}-\sqrt{2}}{4}\\ tan\left(-555^o\right)=\dfrac{sin\left(-555^o\right)}{cos\left(-555^o\right)}=-2+\sqrt{3}\\ cot\left(-555^o\right)=\dfrac{1}{tan\left(-555^o\right)}=\dfrac{1}{-2+\sqrt{3}}=-2-\sqrt{3}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
a) \(log_29\cdot log_34=4\)
b) \(log_{25}\cdot\dfrac{1}{\sqrt{5}}=-\dfrac{1}{4}\)
c) \(log_23\cdot log_9\sqrt{5}\cdot log_54=\dfrac{1}{2}\)
\(log_9\left(\dfrac{1}{27}\right)=log_{3^2}3^{-3}=\dfrac{log_33^{-3}}{log_33^2}=-\dfrac{3}{2}\)