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Bài 1:
a, (-8) . 25. (-2) . 125 . (-5)
= [(-8) . 125] .[ 25 . (-5)]
= (-1000) . (-125)
= 125 000
Bài 1:
a) \(2^8.2.4=2^9.2^2=2^{11}\)
b) \(8^5:64=8^5:8^2=8^3\)
c) \(3^7:9=3^7:3^2=3^5\)
d) \(9^{17}.81=9^{17}.9^2=9^{19}\)
e) \(x^6.x.x^2=x^9\)
Bài 2:
a) \(2^x-15=17\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy x = 5
b) \(2.3^x=162\)
\(3^x=162:2\)
\(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
Vậy x = 4
c) \(5.x.5^2=10\)
\(\Rightarrow x.5^3=10\)
\(\Rightarrow x.125=10\)
\(\Rightarrow x=10:125\)
\(\Rightarrow x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\)
d) \(5.x^2-1=124\)
\(\Rightarrow5.x^2=125\)
\(\Rightarrow x^2=125:5\)
\(\Rightarrow x^2=5^2\)
\(\Rightarrow x=\pm5\)
Vậy \(x=\pm5\)
Câu 1:
a)28.2.4=28.2.22=211
b)85:64=85:82=83
c)37:9=37:32=35
d)917.81=917.92=919
e)x6.x.x2=x9
a, \(^{2^2}\) x- 49 = 5. \(^{3^2}\)
4x - 49 = 45
4x = 45+49
4x = 94
x = 94 :4
x = 26
b, 2\(^{^x}\) : 25 = 1
2\(^{^x}\) 5\(^{^2}\) = 1
2\(^{^x}\) = 2
x = 2 : 2
x = 1
\(2^x.4=128.\)
\(2^x=128:4.\)
\(2^x=32.\)
\(2^x=2^5\Rightarrow x=5\in N.\)
Vậy \(x=5.\)
\(\left(2x+1\right)^3=125.\)
\(\left(2x+1\right)^3=5^3.\)
\(\Rightarrow2x+1=5.\)
\(\Leftrightarrow2x=4.\)
\(\Leftrightarrow x=4:2=2\in N.\)
Vậy \(x=2.\)
a) \(2^x.4=128\)
\(2^x=128:4\)
\(2^x=32\)
\(2^x=2^5\)
\(\Leftrightarrow x=5\left(tm\right)\)
Vậy ...................
b) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(\Leftrightarrow2x+1=5\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy ..............
\(\left(x-2\right)\left(x-4\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2< 0\\x-4>0\end{matrix}\right.=>4< x< 2\left(1\right)\\\left\{{}\begin{matrix}x-2>0\\x-4< 0\end{matrix}\right.=>2< x< 4\left(2\right)}\end{matrix}\right.\)(1 ) vô lý=> loại
=> (x-2).(x-4)<0 <=> 2<x<4
b. ta có\(x^2+1>0\forall x\)
=>(x2 -1).(x2+1)<0 <=> (x2 -1)<0 <=> x2<1
<=> -1<x<1
câu c bạn làm tương tự
câu a
\(\left(2x-2017\right)^2=289\\ < =>\left[\begin{matrix}2x-2017=17\\2x-2017=-17\end{matrix}\right.\\ < =>\left[\begin{matrix}x=1017\left(tm\right)\\x=1000\left(tm\right)\end{matrix}\right.\)
vậy...
câu b
\(\left(\left|x\right|+2016\right)\left(2018-2\left|x\right|\right)=0\\ < =>\left[\begin{matrix}\left|x\right|+2016=0\\2018-2\left|x\right|=0\end{matrix}\right.\\ < =>\left[\begin{matrix}\left[\begin{matrix}x=2016\\x=-2016\end{matrix}\right.\\\left[\begin{matrix}x=1009\\x=-1009\end{matrix}\right.\end{matrix}\right.\) (tm)
vậy ...
câu c
(x - 2016) (2y + 2017) = 5
<=> (x - 2016) (2y + 2017) = 1 . 5 = (-1) (-5)
xét thấy 2y + 2017 là số lẻ
=> \(\left[\begin{matrix}2y+2017=5\\2y+2017=-5\end{matrix}\right.\)
=> \(\left[\begin{matrix}\left\{\begin{matrix}x-2016=1\\2y+2017=5\end{matrix}\right.\\\left\{\begin{matrix}x-2016=-1\\2y+2017=-5\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[\begin{matrix}\left\{\begin{matrix}x=2017\\y=-1006\end{matrix}\right.\\\left\{\begin{matrix}x=2015\\y=-1011\end{matrix}\right.\end{matrix}\right.\) (tm)
vậy ...
số nguyên dương lớn nhất có 3 cs khác nhau là 987
=> lx-2l = 987
<=> x-2 = 987 hoặc x-2 = -987
<=> x=989 hoặc x=-985 (tm)
vậy ...
Bài 1:
\(A=\dfrac{2}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)
\(A=\dfrac{2}{1.5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\right)\)
\(A=\dfrac{2}{5}+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{93}\right)\)
\(A=\dfrac{2}{5}+\dfrac{1}{4}.\dfrac{88}{465}\)
\(A=\dfrac{2}{5}+\dfrac{22}{465}=\dfrac{208}{465}\)
1. Mk sửa lại đề bài như sau:
\(A=\dfrac{1}{1.5}+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{89.93}\)
\(\Rightarrow4A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{89.93}\)
\(4A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{89}-\dfrac{1}{93}\)
\(4A=1-\dfrac{1}{93}\)
\(4A=\dfrac{92}{93}\)
\(A=\dfrac{92}{93}:4\)
\(A=\dfrac{23}{93}\)
2. Mk cux sửa lại đề bài:
\(A=3+3^2+3^3+3^4+3^5+...+3^{100}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(=3\left(1+3+9+27\right)+...+3^{97}\left(1+3+9+27\right)\)
\(=3.40+...+3^{97}.40\)
\(=\left(3+3^{97}\right)⋮4.10\)
\(\Rightarrow A⋮4;10\)
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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
1
a) (-8) . 25 . (-2) . 125 . (-5)=(-8.-2).5^2.25.5.(-5)
= 16.5^2.5^2.(-5) = 4^2.(-5^5)= -50000
b)19 . 25 + 9 . 95 + 19 . 30=19.(25+30)+9.95
= 19.55+9.95 = 19.55+9.(55+40) = 19.55+9.55+9.40= (19+9).55 + 9.40 = 28.55+9.40 = 1900