\(18^{20}.45^5.5^{25}.8^{10}\)

\(=3^{40}.2^{20}.5^5.3^{10}.5^{25}.2^{30}\)

\(=3^{50}.2^{50}.5^{30}\)

\(=6^{50}.5^{30}\) 

\(=\left(6^5\right)^{10}.\left(5^3\right)^{10}\)

\(=\left(6^5.5^3\right)^{10}\)

\(\left(x^2y\right)^5.\left(x^2.y^2\right)^7.\left(x.y\right)^6.x^3\)

\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^3.x^3\)

\(=x^{33}.y^{22}\)

\(=\left(x^3\right)^{11}.\left(y^2\right)^{11}\)

\(=\left(x^3.y^2\right)^{11}\)

\(2^7.3^8.4^9.9^8\)

\(=2^7.3^8.2^{18}.3^{16}\)

\(=2^{25}.3^{24}\)( mk chỉ làm được đến thế thôi )

Tham khảo nhé~

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 )

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)\(2^7.3^8.4^9.9^8=2^7.3^8.2^{18}.3^{16}=2^{25}.3^{24}\)

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)

b) \(5xy^2-10xyz+5xz^2\)

\(=5xy^2-5xyz-5xyz+5xz^2\)

\(=5xy\left(y-z\right)-5xz\left(y-z\right)\)

\(=\left(y-z\right)\left(5xy-5xz\right)\)

\(=5x\left(y-z\right)\left(y-z\right)\)

\(=5x\left(y-z\right)^2\)

c) \(3x^2\left(a+b+c\right)+36xy\left(a+b+c\right)+108y^2\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(3x^2+36xy+108y^2\right)\)

\(=\left(a+b+c\right)\left(\sqrt{3x}+6\sqrt{3y}\right)^2\)

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!!!!

a.\(\frac{5}{6}-\left|1-x\right|=\frac{1}{2}\)

\(=>\left|1-x\right|=\frac{5}{6}-\frac{1}{2}\)

\(=>\left|1-x\right|=\frac{1}{3}\)

\(=>\orbr{\begin{cases}1-x=\frac{1}{3}\\1-x=\frac{-1}{3}\end{cases}}\)

\(=>\orbr{\begin{cases}x=1-\frac{1}{3}=\frac{2}{3}\\x=1-\frac{-1}{3}=\frac{4}{3}\end{cases}}\)

`a, (x-y)^2 = (x+y)^2 - 4xy = 12^2 - 35 . 4 = 144 - 140 = 4`.

`b, (x+y)^2 = (x-y)^2 + 4xy = 8^2 + 20.4 = 64 + 80 = 144`

`c, x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 5^3 - 3 . 6 . 5 = 125 - 90 = 35`

`d, x^3 - y^3 = (x-y)^3 - 3xy(x-y) = 3^3 - 3 .40 . 3 = 27 - 360 = -333`.

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)\(^{12}\)

\(x:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^2\)

\(x=\left(\dfrac{3}{4}\right)^2.\left(\dfrac{3}{4}\right)^3\) <=> \(x=\left(\dfrac{3}{4}\right)^{2+3}\)

=> \(x=\left(\dfrac{3}{4}\right)^5\)

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: \(\left(x^m\right)^n=x^{m.n}\)

^{\(0,36^8=\left[\left(0,6\right)^2\right]^8=\left(0,6\right)^{16}\)}

\(0,216^4=\left[\left(0,6\right)^3\right]^4=\left(0,6\right)^{12}\)

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: \(-\dfrac{3}{2}\le x\le\dfrac{7}{3}\)

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: \(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)

\(\Rightarrow\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}=2\)

\(\Rightarrow\left\{{}\begin{matrix}1\le x\le3\\y=-3\end{matrix}\right.\)

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