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Lời giải:
Giả sử \(\log _{3}a=\log_4b=\log_{12}c=\log_{13}(a+b+c)=t\)
\(\Rightarrow 13^t=3^t+4^t+12^t\)
\(\Rightarrow \left ( \frac{3}{13} \right )^t+\left ( \frac{4}{13} \right )^t+\left ( \frac{12}{13} \right )^t=1\)
Xét vế trái , đạo hàm ta thấy hàm luôn nghịch biến nên phương trình có duy nhất một nghiệm \(t=2\)
Khi đó \(\log_{abc}144=\log_{144^t}144=\frac{1}{t}=\frac{1}{2}\)
Đáp án B
cho em hỏi tại sao lại có 3^t +4^t +12^t=13^t. Với lại em không hiểu chỗ tại sao hàm số nghịch biến. Và tại sao từ \(\log_{abc}144=\log144_{144^t}=\dfrac{1}{t}\)
a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)
\(=2^3+30-\dfrac{3}{2}\)
\(=36,5\)
b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)
\(=0,1^{-1}-2^2-2^{-4}\)
\(=10-4-\dfrac{1}{16}\)
\(=\dfrac{95}{16}\)
\(\Leftrightarrow\left(1-2i\right)z-\left(\dfrac{1}{2}-\dfrac{3}{2}i\right)=\left(3-i\right)z\)
\(\Leftrightarrow\left(1-2i\right)z-\left(3-i\right)z=\dfrac{1}{2}-\dfrac{3}{2}i\)
\(\Leftrightarrow\left(-2-i\right)z=\dfrac{1}{2}-\dfrac{3}{2}i\)
\(\Rightarrow z=\dfrac{1-3i}{2\left(-2-i\right)}=\dfrac{1}{10}+\dfrac{7}{10}i\)
\(\Rightarrow A\left(\dfrac{1}{10};\dfrac{7}{10}\right)\) \(\Rightarrow\) tọa độ trung điểm I là \(\left(\dfrac{1}{20};\dfrac{7}{20}\right)\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
a) \(2^{-2}=\dfrac{1}{2^2}< 1\)
b) \(\left(0,013\right)^{-1}=\dfrac{1}{0,013}>1\)
c) \(\left(\dfrac{2}{7}\right)^5=\dfrac{2^5}{7^5}< 1\)
d) \(\left(\dfrac{1}{2}\right)^{\sqrt{3}}=\dfrac{1}{2^{\sqrt{3}}}< \dfrac{1}{2^{\sqrt{1}}}=\dfrac{1}{2}< 1\)
e) vì \(0< \dfrac{\pi}{4}< 1\)
Suy ra \(\left(\dfrac{\pi}{4}\right)^{\sqrt{5}-2}=\dfrac{\left(\dfrac{\pi}{4}\right)^{\sqrt{5}}}{\left(\dfrac{\pi}{2}\right)^2}>\dfrac{\left(\dfrac{\pi}{4}\right)^{\sqrt{4}}}{\left(\dfrac{\pi}{4}\right)^2}=1\)
f) Vì \(0< \dfrac{1}{3}< 1\)
Nên \(\left(\dfrac{1}{3}\right)^{\sqrt{8}-3}>\left(\dfrac{1}{3}\right)^{\sqrt{9}-3}=\left(\dfrac{1}{3}\right)^0=1\)
\(\left(\dfrac{-2}{5}+\dfrac{3}{7}\right)-\left(\dfrac{4}{9}+\dfrac{12}{20}-\dfrac{13}{35}\right)+\dfrac{7}{35}\)
\(=-\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{4}{9}-\dfrac{3}{5}+\dfrac{13}{35}+\dfrac{7}{35}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{13}{35}+\dfrac{7}{35}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+1-\dfrac{4}{9}\\ =-\dfrac{4}{9}\)