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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 \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)
a) \(4.5^2-32:2^5\)
\(=4.25-2^5:2^5\)
\(=100-1\)
\(=99.\)
b) \(9.8.14+6.\left(-17\right)\left(-12\right)+19.\left(-4\right).18\)
\(=9.2.4.14+6.3.\left(-4\right)\left(-17\right)+76.18\)
\(=18.56+18.68+18.76\)
\(=18\left(56+68+76\right)\)
\(=18\left(132+68\right)\)
\(=18.200\)
\(=3600.\)
c) \(\left(\dfrac{-1}{2}\right)^3-2.\left(\dfrac{-1}{2}\right)^2+3.\left(\dfrac{-1}{2}\right)+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\left(\dfrac{-1}{2}\right)^2+2.\dfrac{-1}{2}+3\right]+1\)
\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+\left(-1\right)+3\right]+1\)
\(\)\(=\left(\dfrac{-1}{2}\right)\left[\dfrac{1}{4}+2\right]+1\)
\(=\left(\dfrac{-1}{2}\right).\dfrac{9}{4}+1\)
\(=\dfrac{-9}{8}+1\)
\(=\dfrac{-1}{8}\)
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 }
\(A=64^{11}\cdot16^{13}=2^{66}\cdot2^{52}=2^{118}\)
\(B=32^{17}\cdot8^{19}=2^{85}\cdot2^{57}=2^{142}\)
Do đó: A<B