5*5 mũ x=62
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`43 + 338:(6x+5) = 5^2. 2`
`<=> 43 + 338:(6x+5) = 50`.
`<=> 338:(6x+5) = 7`.
`<=> 6x + 5 = 338 :7`
`<=> 6x + 5= 338/7`.
`<=> 6x = 308/7`
`<=> x = 169/21`.
Vậy `x = 169/21`.
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\(a)9.4.25.8.125\)
\(=9.\left(4.25\right).\left(8.125\right)\)
\(=9.100.1000\)
\(=900.1000\)
\(=900000\)
\(a,9.4.25.8.125=9.\left(4.25\right).\left(8.125\right)=9.100.1000=900000\)
\(b,25.5.27.2\)(câu này hình như thiếu đề, ko tính nhanh đc)
\(c,37.7+80.3+43.7\)\(=7.\left(37+43\right)+80.3\)\(=7.80+3.80=80\left(7+3\right)=80.10=800\)
\(d,113.38+113.62+87.62+87.38\)\(=113\left(38+62\right)+87\left(62+38\right)\)
\(=113.100+87.100=100.\left(113+87\right)=100.200=20000\)
Viết gọn lũy thừa:
\(a,x.x.y.y.x.y.x=\left(x.x.x.x\right).\left(y.y.y\right)=x^4.y^4\)
\(b,7^5:343=7^5:7^3=7^{5-3}=7^2\)
\(c,A^{12}:A^{18}=A^{12-18}=A^{-6}\)
\(d,x^7.x^4.x=x^{7+4+1}=x^{12}\)
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a = 2 + 22 +23+........................+ 2100 chia hết cho 62
a = [ 2 + 22 +23+.24+25 ] +[ 26 +27 +28+29+210 ] + ...........+ [ 296 + 297 +298 +299 + 2100 ]
a= 62 + [ 210 . 62 ] + [ 215 . 62 ] + [ 220. 62 ] + ......................+ [ 2100 . 62 ]
a= 62 . [ 210 + 215 + 220 +......................+ 2100 ]
Mà 62 chia hết cho 62 => 62 . [ 210 + 215 + 220 +......................+ 2100 ] hay a chia hết cho 62
a = (2+2^2+2^3+2^4+2^5)+(2^6+2^7+2^8+2^9+2^10)+.....+(2^96+2^97+2^98+2^99+2^100)
= 62+2^5.(2+2^2+2^3+2^4+2^5)+......+2^95.(2+2^2+2^3+2^4+2^5)
= 62+2^5.62+....+2^95.62
= 62.(1+2^5+....+2^95) chia hết cho 62
=> ĐPCM
k mk nha
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a: 14/35=2/5
b: 23/62=23/62
c: -33/143=-3/11
d: -53/81=-53/81
e: \(\dfrac{2^2\cdot5^2}{2^5\cdot5^4}=\dfrac{1}{2^3\cdot5^2}=\dfrac{1}{8\cdot25}=\dfrac{1}{200}\)
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a) \(\dfrac{10^{12}+5^{11}.2^9-5^{13}.2^8}{4.5^5.10^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^2.5^5.2^6.5^6}\)
\(=\dfrac{2^{12}.5^{12}+5^{11}.2^9-5^{13}.2^8}{2^8.5^{11}}\)
\(=\dfrac{\left(2^8.5^{11}\right)\left(2^4.5+2-5^2\right)}{2^8.5^{11}}\)
\(=2^4.5+2-5^2\)
\(=57\)
b) \(\dfrac{\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y-x\right)^2}\)
\(=\dfrac{\left(x^2+y^2-2xy\right)\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]}{\left(y^2+x^2-2xy\right)}\)
\(=5\left(x-y\right)^2-3\left(x-y\right)+4\)
c) \(\dfrac{\left(x+y\right)^5-2\left(x+y\right)^4+3\left(x+y\right)^3}{-5\left(x+y\right)^3}\)
\(=\dfrac{\left(x+y\right)^3\left[5\left(x+y\right)^2-2\left(x+y\right)+3\right]}{-5\left(x+y\right)^3}\)
\(=\dfrac{5\left(x+y\right)^2-2\left(x+y\right)+3}{-5}\)
sai đề rồi