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a)\(10^{19}+10^{18}+10^{17}=10^{17}\left(10^2+10+1\right)\)=1017.111=1016.2.5.111=1016.2.555 chia hết cho 555
b)\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)=328-327-326=325(33-32-3)=325.15 chia hết cho 15
c)\(5^7-5^6+5^5=5^5\left(5^2-5+1\right)=5^5.21\) chia hết cho 21
d)\(7^6+7^5-7^4=7^3\left(7^3+7^2-7\right)=7^3.385=7^3.5.77\) chia hết cho 77
1)a)34-26-54=81-64-625=-608
ko hiểu b
2)a)(3x-1)2=(5/6)2=(-5/6)2
+)3x-1=5/6 =>x=11/18
+)3x-1=-5/6 =>x=1/18
b)(x+7)x-11(1-(x-7)23)=0
=>+)(x+7)x-11=0 =>x+7=0 =>x=-7
+)1-(x+7)23=0 =>(x+7)23=1 =>x+7=1 =>x=-6
\(1.a)\) Ta có: \(\left\{{}\begin{matrix}64^8=\left(8^2\right)^8=8^{16}\\16^{12}=8^{12}.2^{12}=8^{12}.\left(2^3\right)^4=8^{12}.8^4=8^{16}\end{matrix}\right.\)
Có: \(8^{16}=8^{16}\Rightarrow64^8=16^{12}\)
Vậy...
\(b)\) Ta có: \(\left\{{}\begin{matrix}\left(-5\right)^{30}=\left[\left(-5\right)^3\right]^{10}=\left(-125\right)^{10}\\\left(-3\right)^{50}=\left[\left(-3\right)^5\right]^{10}=\left(-243\right)^{10}\end{matrix}\right.\)
Có: \(\left(-125\right)^{10}< \left(-243\right)^{10}\Rightarrow\left(-5\right)^{30}< \left(-3\right)^{50}\)
Vậy...
\(c)\) Ta có: \(\left\{{}\begin{matrix}2^{27}=\left(2^3\right)^9=8^9\\3^{18}=\left(3^2\right)^9=9^9\end{matrix}\right.\)
Có: \(8^9< 9^9\Rightarrow2^{27}< 3^{18}\)
Vậy...
\(d)\) Ta có: \(\left\{{}\begin{matrix}\left(\dfrac{1}{25}\right)^{10}=\left[\left(\dfrac{1}{5}\right)^2\right]^{10}=\left(\dfrac{1}{5}\right)^{20}\\\left(\dfrac{1}{125}\right)^8=\left[\left(\dfrac{1}{5}\right)^3\right]^8=\left(\dfrac{1}{5}\right)^{24}\end{matrix}\right.\)
Có: \(\left(\dfrac{1}{5}\right)^{20}< \left(\dfrac{1}{5}\right)^{24}\Rightarrow\left(\dfrac{1}{24}\right)^{10}< \left(\dfrac{1}{125}\right)^8\)
Vậy...
\(e)\)Có: \(32^9=\left(2^5\right)^9=2^{45}< 2^{52}=\left(2^4\right)^{13}=16^{13}< 18^{13}\)
\(\Rightarrow32^9< 18^{13}\)
Vậy...