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a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)
\(\dfrac{6^6+2^7.3^6+2^6.3^7}{4.2^3.3^2.2^2}\)
\(=\dfrac{\left(2^2.3\right)^6+2^7.3^6+2^6.3^7}{2^2.2^3.3^2.2^2}\)\(=\dfrac{2^{12}.3^6+2^7.3^6+2^6.3^7}{2^7.3^2}\)
\(=\dfrac{2^6.3^6.\left(2^6+2+3\right)}{2^7.3^2}\)\(=\dfrac{3^4.69}{2}=\dfrac{5589}{2}\)
a, \(3^{-2}.3^4.3^x=3^7\)
\(\Rightarrow3^{-2+4+x}=3^7\)
\(\Rightarrow3^{2+x}=3^7\)
Vì \(3\ne\pm1;3\ne0\) nên \(2+x=7\Rightarrow x=5\)
b, \(2^{-1}.2^x+4.2^x=9.2^5\)
\(\Rightarrow2^x\left(2^{-1}+4\right)=288\)
\(\Rightarrow2^x.4,5=288\Rightarrow2^x=64=2^6\)
Vì \(2\ne\pm1;2\ne0\) nên \(x=6\)
Chúc bạn học tốt!!!
a: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
b: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(-2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
giups mình với mai mình thi rồi!
đề bài là tính hả bn?