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\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
= \(\dfrac{2}{2}.\left(\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{59.61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
= \(\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
=\(\dfrac{3}{2}.\dfrac{56}{305}\)
= \(\dfrac{78}{305}\)
\(\left(x^2-4\right)\left(6-2x\right)=0\) ⇔ \(x^2-4=0\) 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)410. 815 = (22)10.(23)15
=220.245
=265
b) 415 . 530 = 415 . (52)15
= 415 . 2515
= (4.25)15
= 10015
c) 2716 : 910 = (33)16 : (32)10
=348 : 320
=328
d)A=\(\frac{72^3.54^2}{108^4}\)
TS:723.542=(2.2.2.3.3)3.(2.3.3.3)2
=23.23.23.33.33.22.32.32.32
=23+3+3+2.33+3+2+2+2
=211.312
MS:1084=(2.2.3.3.3)4
=24. 24.34.34.34
=24+4 . 34+4+4
=28 . 312
\(\Rightarrow\)A=\(\frac{2^{11}.3^{12}}{2^8.3^{12}}\) = 23=8
e)B=\(\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
TS:310 .11 + 310 .5 = 310 . (11+5)
= 310 . 16
= 310. 24
\(\Rightarrow\)B=\(\frac{3^{10}.2^4}{3^9.2^4}\) = 31=3
Giải:A=723.542/(54.2)4=723.542/544.24=723/542.24=35.26/35.26=1. B=39.3.11+39.3.5/39.24=39.(33+15)/39.24=48/16=3. C=210.(13+65)/22.104=28.78/104=28.26.3/26.4=28.3/4=26.3=192. a)->2x=128:4=32.=>x=5(25=32) b)->2x+1=5(53=125)=>x=2. c)Ko có số nào ngoài 1 và 0 tồn tại dưới dạng(x-5)4=(x-5)6 ->Nếu x-5=0=>x=5 ->Nếu x-5=1=>x=6
0
( 2^3 x 3^2)^3 x ( 2 x 3^3 )^2 = 2^9 x 3^6 x 2^2 x 3^6 = 2^11 x 3^12
( 2^2 x 3^3 ) = 2^8 x 3^12 = 2^8 x 3^12
= 2^3 = 8
bài 8
c) chứng minh \(\overline{aaa}⋮37\)
ta có: \(aaa=a\cdot111\)
\(=a\cdot37\cdot3⋮37\)
\(\Rightarrow aaa⋮37\)
k mk nha
k mk nha.
#mon
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a, (0,25)3.32
= 0,5
b, \(\dfrac{72^3.54^2}{108^4}=\dfrac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}\)\(=\dfrac{2^9.3^6.2^2.3^6}{2^8.3^{12}}\)
\(=\dfrac{2^{11}.3^{12}}{2^8.3^{12}}=2^3\)
c, \(\dfrac{81^{11}.3^{17}}{27^{10}.9^{15}}=\dfrac{\left(3^4\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\dfrac{3^{44}.3^{17}}{3^{30}.3^{30}}\)\(=\dfrac{3^{61}}{3^{60}}=3\)
@Lớp 6B Đoàn Kết