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

\(=\left(2.3^2\right)^{20}.\left(3^2.5\right)^5.5^{25}.\left(2^3\right)^{10}\)

\(=2^{20}.3^{40}.3^{10}.5^5.5^{25}.2^{30}\)

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

\(=6^{50}.5^{30}\)

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

\(=7776^{10}.125^{10}\)

\(=972000^{10}\)

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

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

\(=x^{33}.y^{25}\)

\(=x^{25}.y^{25}.x^8\left(?\right)\)

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

\(=2^7.3^8.\left(2^2\right)^9.\left(3^2\right)^8\)

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

\(=2^{25}.3^{24}\)\(\left(?\right)\)

Bạn trả lời đúng rồi a, Mình tick cho bạn nha

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

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

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

\(=6^{50}.5^{30}\) 

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

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

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

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

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

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

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

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

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

\(=2^{25}.3^{24}\)( 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,ta có:

\(3^5.5^7.5.3^2\)

\(=\left(3^5.3^2\right).\left(5.5^7\right)\)

\(=3^7.5^8\)

\(b=2^8.\left(2.2\right)^5.9^9\)

\(=2^8.2^{10}.9^9\)

\(=2^{18}.9^9\)

Vì bài dài nên mình sẽ tách ra nhé.

1a. Ta có:

$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$

$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$

$=-3(-z)(-x)(-y)=3xyz$

$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$

------------------------

$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$

$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$

$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$

$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$

$=-z^5+5xyz^3-5x^2y^2z$

$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$

$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$

Từ $(1);(2)$ ta có đpcm.

1b.

$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$

$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$

$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$

Do đó:

$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$

$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$

$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$

$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$

$=7xyz(x^2y^2-2xyz^2+z^4)$

$=7xyz(xy-z^2)$

$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$

$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$

$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)

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

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

a)\(12^3.3^{-4}.64\)

\(=3^3.2^6.3^{-3}.2^6=2^{12}\)

b) \(\left(\frac{3}{7}\right)^5.\left(\frac{7}{3}\right)^{-1}.\left(\frac{5}{3}\right)^6:\left(\frac{343}{625}\right)^2\)

\(=\frac{3^5.7^{-1}}{7^5.3^{-1}}.\left(\frac{5}{3}\right)^6:\frac{7^6}{5^8}\)

\(=\frac{3^6}{7^6.}.\frac{5^6}{3^6}.\frac{5^8}{7^6}\)

\(=\frac{5^{14}}{7^{12}}\)

( Mik làm mấy phần mà bạn dưới chưa làm)

11) xy+x+y=9

\(\Leftrightarrow\) xy+x+y+1=9+1

\(\Leftrightarrow\left(xy+x\right)+\left(y+1\right)\)=10

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=10\)

\(\Leftrightarrow\) (x+1)(y+1)=10=1.10=10.1=-1.-10=-10.-1=2.5=5.2=-2.-5=-5.-2

\(\Rightarrow\) TH1: x+1=1 ; y+1=10

\(\Leftrightarrow x=0;y=9\)

TH2: x+1=10;y+1=1

\(\Leftrightarrow\)x=9;y=0

TH3: x+1=-1;y+1=-10

\(\Leftrightarrow\) x=-2;y=-11

...........

Vậy:........

( Bạn tự làm nốt chứ dài quá, mik chỉ hướng dẫn cách làm bài thôi)

1) -x = -7

=> x = 7

2) - x = 17

=> x = - 17

3) |x| = 17

=> x = ±17

4) -(-x) = |-17|

=> x = 17

5) - 19 - x = 17

=> - x = 17 + 19

=> x = - 36

6) - 19 - x = - 17

=> - x = - 17 + 19

=> -x = 2

=> x = - 2

7) - 5 - (10 - x) = 7

=> - 5 - 10 + x = 7

=> - 15 + x = 7

=> x = 7 + 15

=> x = 22

8) |x + 3| + 7 = 12

=> |x + 3| = 12 - 7

=> |x + 3| = 5

=> x + 3 = 5 hoặc x + 3 =- 5

=> x = 2 hoặc x = - 8

9) 2 - |x - 2| = x

=> - |x - 2| - x = - 2

TH1: x >= 2

- (x - 2) - x = - 2

=> - x + 2 - x =- 2

=> - 2x = - 4

=> x = 2 (nhận)

TH2: x < 2

-[-(x - 2)] - x = - 2

=> x - 2 - x = - 2

=> 0x = 0 (vô số nghiệm)

...

\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)

\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)