cho hình vẽ,biết Ax//By;góc a=40 độ;góc AOB=80 độ
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\(\dfrac{9^{15}\cdot8^{11}}{3^{29}\cdot16^8}=\dfrac{3^{2^{15}}\cdot2^{3^{11}}}{3^{29}\cdot2^{4^8}}=\dfrac{3^{30}\cdot2^{33}}{3^{29}\cdot2^{32}}\)
\(\Rightarrow\dfrac{3^{30}\cdot2^{33}}{3^{29}\cdot2^{32}}=\dfrac{3^{29}\cdot2^{32}\cdot3\cdot2}{3^{29}\cdot2^{32}}=3\cdot2=6\)
\(2^x+2^{x+3}=144\)
\(\Leftrightarrow2^x\left(1+2^3\right)=144\)
\(\Leftrightarrow2^x.9=144\)
\(\Leftrightarrow2^x=144:9\)
\(\Leftrightarrow2^x=16\)
\(\Leftrightarrow2^x=2^4\)
\(\Leftrightarrow x=4\)
\(\left[{}\begin{matrix}3x+\sqrt{2}=4\\3x+\sqrt{2}=-4\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}3x=4-\sqrt{2}\\3x=-4-\sqrt{2}\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=\dfrac{4-\sqrt{2}}{3}\\x=\dfrac{-4-\sqrt{2}}{3}\end{matrix}\right.\)
\(7^{2x}+7^{2x+3}=344\)
\(7^{2x}+7^{2x}.7^3=344\)
\(7^{2x}+7^{2x}.343=344\)
\(7^{2x}.1+7^{2x}.343=344\)
\(7^{2x}.\left(1+343\right)=344\)
\(7^{2x}.344=344\)
\(7^{2x}=344:344\)
\(7^{2x}=1\)
\(\Rightarrow7^{2x}=7^0\)
\(2x=0\)
\(x=0:2\)
\(x=0\)