Viết các tích dưới dạng lũy thừa :
a) \(18^{20}.45^5.5^{25}.8^{10}\)
b) \(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
c) \(2^7.3^8.4^9.9^8\)
K
Khách
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23 tháng 7 2018
\(18^{20}.45^5.5^{25}.8^{10}\)
\(=3^{40}.2^{20}.5^5.3^{10}.5^{25}.2^{30}\)
\(=3^{50}.2^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=\left(6^5.5^3\right)^{10}\)
\(\left(x^2y\right)^5.\left(x^2.y^2\right)^7.\left(x.y\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^3.x^3\)
\(=x^{33}.y^{22}\)
\(=\left(x^3\right)^{11}.\left(y^2\right)^{11}\)
\(=\left(x^3.y^2\right)^{11}\)
\(2^7.3^8.4^9.9^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)( mk chỉ làm được đến thế thôi )
Tham khảo nhé~
23 tháng 7 2018
a) \(18^{20}.45^5.5^{25}.8^{10}\)
\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)
\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=7776^{10}.125^{10}\)
\(=972000^{10}\)
b ) \(\left(x^2y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^6.x^3\)
\(=x^{33}.y^{25}\)
\(=x^{25}.y^{25}.x^8\)
\(=...\)
c) \(2^7.3^8.4^9.9^8\)
\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)
\(=...\)( Câu c này hình như đề bài sai sót . Không chuyển thành lũy thừa được )
T
23 tháng 7 2018
a)\(3^5.5^7.45=3^5.5^7.3^2.5=3^7.5^8\)
b)\(2^8.4^5.9^9\)\(=2^8.2^{10}.9^9=2^{18}.9^9\)
c)\(\left(2^3.3^5.5^7\right)^{10}.12^{20}=2^{13}.3^{15}.5^{17}.12^{20}\)\(=2^{13}.3^{15}.5^{17}.2^{40}.3^{20}=2^{53}.3^{35}.5^{17}\)
d)\(\left(x^2y\right)^5.\left(x^2y^2\right)^7.\left(x.y^2\right)^6.x^3=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
e)\(18^{20}.45^5.5^{25}.8^{10}=2^{20}.3^{40}.5^5.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{35}=6^{50}.5^{35}\)
f)\(2^7.3^8.4^9.9^8=2^7.3^8.2^{18}.3^{16}=2^{25}.3^{24}\)
OT
14 tháng 8 2019
a)ta co: 125x^3+y^6=(5x)^3+(y^2)^3=(5x+y^2)(5x-5xy^2+y^2) b)ta co 5xy^2-10xyz+5xz^2=5x(y^2-2yz+z^2)=5x(y-z)^2 (may cau sau gan giong ban tu lam nha)
15 tháng 8 2019
b) \(5xy^2-10xyz+5xz^2\)
\(=5xy^2-5xyz-5xyz+5xz^2\)
\(=5xy\left(y-z\right)-5xz\left(y-z\right)\)
\(=\left(y-z\right)\left(5xy-5xz\right)\)
\(=5x\left(y-z\right)\left(y-z\right)\)
\(=5x\left(y-z\right)^2\)
28 tháng 5 2020
a) \(\frac{5}{6}-\left|1-x\right|=\frac{1}{2}\)
<=> \(\left|1-x\right|=\frac{5}{6}-\frac{1}{2}\)
<=>\(\left|1-x\right|=\frac{1}{3}\)
<=>\(\orbr{\begin{cases}1-x=\frac{1}{3}\\1-x=-\frac{1}{3}\end{cases}}\)
<=>\(\orbr{\begin{cases}x=1-\frac{1}{3}\\x=1-\left(-\frac{1}{3}\right)\end{cases}}\)
<=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{4}{3}\end{cases}}\)
Vậy \(x=\frac{2}{3}\)hoặc\(x=\frac{4}{3}\)
b) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.\left(-10\right)^9}{4^{10}.5^7+20^8.4}.\frac{144}{25}\)
\(=\frac{2^{15}.5^8+2^5.2^9.5^9}{2^{20}.5^7+2^{16}.5^8.2^2}.\frac{2^4.3^2}{5^2}\)
\(=\frac{2^{14}.5^8.\left(2+5\right)}{2^{18}.5^7.\left(2^2+5\right)}.\frac{2^4.3^2}{5^2}\)
\(=\frac{2^{14}.5^8.7}{2^{18}.5^7.3^2}.\frac{2^4.3^2}{5^2}\)
\(=\frac{2^{18}.3^2.5^8.7}{2^{18}.3^2.5^9}\)
\(=\frac{7}{5}\)
Học tốt!!!!
JB
28 tháng 5 2020
a.\(\frac{5}{6}-\left|1-x\right|=\frac{1}{2}\)
\(=>\left|1-x\right|=\frac{5}{6}-\frac{1}{2}\)
\(=>\left|1-x\right|=\frac{1}{3}\)
\(=>\orbr{\begin{cases}1-x=\frac{1}{3}\\1-x=\frac{-1}{3}\end{cases}}\)
\(=>\orbr{\begin{cases}x=1-\frac{1}{3}=\frac{2}{3}\\x=1-\frac{-1}{3}=\frac{4}{3}\end{cases}}\)
16 tháng 9 2018
1.
a) x : \(\left(\dfrac{3}{4}\right)^3\) =\(\left(\dfrac{3}{4}\right)^3\)
x = \(\left(\dfrac{3}{4}\right)^3.\left(\dfrac{3}{4}\right)^3\)
x = \(\dfrac{3}{4}^{3+3}\)
x = \(\dfrac{3}{4}^6\)
x = \(\dfrac{729}{4096}\)
b) \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
x = \(\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\)
x = \(\dfrac{2}{5}^{8-5}\)
x = \(\dfrac{2}{5}^3\)
x = \(\dfrac{8}{5}\)
2.
(0,36)\(^8\) \([\left(0,6\right)^3]^8\) = (0,6)\(^{3.8}\) = ( 0,6)\(^{24}\)
( 0,216)\(^4\) = \([\left(0,6\right)^3]^4\) = (0.6)\(^{3.4}\) = ( 0,6)\(^{12}\)
M
18 tháng 9 2018
\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)
\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)
=> \(x=\left(\dfrac{3}{4}\right)^5\)
b, \(\left(\dfrac{2}{5}\right)^5.x=\left(\dfrac{2}{5}\right)^8\)
\(x=\left(\dfrac{2}{5}\right)^8:\left(\dfrac{2}{5}\right)^5\Leftrightarrow x=\left(\dfrac{2}{5}\right)^{8-5}\)
=>\(x=\left(\dfrac{2}{5}\right)^3\)
bài 2 : Với bài này ta cần áp dụng quy tắc: \(\left(x^m\right)^n=x^{m.n}\)
\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)
\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)
7 tháng 12 2017
1) a) \(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\ge0\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}7x=5y\\2z=3y\\xy+yz+xz=2000\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}y\\z=\dfrac{3}{2}y\\xy+yz+xz=2000\end{matrix}\right.\)
Ta có: \(xy+yz+xz=2000\)
\(\Rightarrow\dfrac{5}{7}y^2+\dfrac{3}{2}y^2+\dfrac{15}{14}y^2=2000\)
\(\Rightarrow y^2\left(\dfrac{5}{7}+\dfrac{3}{2}+\dfrac{15}{14}\right)=2000\Leftrightarrow\dfrac{23}{7}y^2=2000\)
Tìm \(y\) và suy ra \(x;z\) là được,Bài này nghiệm khá xấu
b) \(\left|3x-7\right|+\left|3x+2\right|+8=\left|7-3x\right|+\left|3x+2\right|+8\ge\left|7-3x+3x+2\right|+8\ge9+8=17\)Dấu "=" xảy ra khi: \(-\dfrac{3}{2}\le x\le\dfrac{7}{3}\)
7 tháng 12 2017
2) a)Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(\Rightarrow\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\left|y+3\right|+5\ge5\\\dfrac{10}{\left(2x-6\right)^2+2}\le\dfrac{10}{2}=5\end{matrix}\right.\)
Mà theo đề bài: \(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(\Rightarrow\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}=5\)
\(\Rightarrow\left\{{}\begin{matrix}y=-3\\x=3\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\\\dfrac{6}{\left|y+3\right|+3}\le\dfrac{6}{3}=2\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)
\(\Rightarrow\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}=2\)
\(\Rightarrow\left\{{}\begin{matrix}1\le x\le3\\y=-3\end{matrix}\right.\)
16 tháng 4 2017
a) Ta thấy: có 5 thừa số (-5) nên tích mang dấu "-" nên:
(-5).(-5).(-5).(-5).(-5) = -55
b) (-2).(-2).(-2).(-3).(-3).(-3)
= (-2).(-3).(-2).(-3).(-2).(-3)
=6.6.6 = 63
hoặc: ta thấy tích có 6 thừa số nguyên âm nên tích mang dấu "+"
(-2).(-2).(-2).(-3).(-3).(-3)
= 23.33
\(a)18^{20}.45^5.5^{25}.8^{10}\)
\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)
\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)
\(=2^{50}.3^{50}.5^{30}\)
\(=6^{50}.5^{30}\)
\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)
\(=7776^{10}.125^{10}\)
\(=972000^{10}\)
\(b)\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(xy\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^6.x^3\)
\(=x^{33}.y^{25}\)
\(=x^{25}.y^{25}.x^8\left(?\right)\)
\(c)2^7.3^8.4^9.9^8\)
\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)
\(=2^7.3^8.2^{18}.3^{16}\)
\(=2^{25}.3^{24}\)\(\left(?\right)\)
Bạn trả lời đúng rồi a, Mình tick cho bạn nha