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b)Có \(63^7< 64^7\)
\(64^7=\left(2^6\right)^7=2^{42}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
Mà \(2^{42}< 2^{48}\Rightarrow63^7< 64^7< 16^{12}\Rightarrow63^7< 16^{12}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) hoặc \(6-2x=0\)
*Nếu \(x^2-4=0\)
⇒ x2 = 4
⇒ x ∈ {2 ; -2}
*Nếu \(6-2x=0\)
⇒2x = 6
⇒ x = 6 : 2 = 3
Vậy x ∈ { -2 ; 2 ; 3 }
a) Ta có : 3111 < 3211 = (25)11 = 255
1714>1614 = (24)14=256
=> 3111 <255<256<1714
=>3111<1714
b)Ta có : 1617 = (24)17 = 268
822 = (23)22 = 266
Vì 268>266 nên 1617 >822
c) Ta có : 10750 <10850= (4.27)50 = 450 .2750 = 2100 . 3150
7375 >7275 = (8.9)75 = 875 . 975 = 2225 . 3150
=> 10750 <2100 .3150 <2225.3150<7375
=> 10750 <7375
d) Ta có : 291 >290 = (25)18 = 3218
535<536 = (52)18 = 2518
Vì 3218 >2518 nên 291 > 535.
e) Ta có : \(\left(\frac{1}{32}\right)^7=\frac{1}{32^7}=\frac{1}{2^{35}}\)
\(\left(\frac{1}{16}\right)^9=\frac{1}{16^9}=\frac{1}{2^{36}}\)
Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
Ta có :
\(A=1+5+5^2+...+5^{32}\)
\(A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{30}+5^{31}+5^{32}\right)\)
\(A=31+5^3\left(1+5+5^2\right)+...+5^{30}\left(1+5+5^2\right)\)
\(A=31+31.5^3+...+31.5^{30}\)
\(A=31\left(1+5^3+...+5^{30}\right)\) chia hết cho 31
Vậy \(A\) chia hết cho 31
\(a)\) Ta có :
\(\frac{a}{b}< \frac{a+c}{b+c}\)
\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow\)\(ab+ac< ab+bc\)
\(\Leftrightarrow\)\(ac< bc\)
\(\Leftrightarrow\)\(a< b\)
Mà \(a< b\) \(\Rightarrow\) \(\frac{a}{b}< 1\)
Vậy ...
Bệnh!!!