C = \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)

= \(\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)

= \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)

= \(\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)

= \(\dfrac{33-9}{4}\)

= 6
@trần phúc nguyên

B = \(\dfrac{45^3.20^4.18^2}{180^5}\)

= \(\dfrac{\left(3^2.5\right)^3\left(2^2.5\right)^4.\left(2.3^2\right)^2}{\left(2^2.3^2.5\right)^5}\)

= \(\dfrac{3^6.5^3.2^8.5^4.2^2.3^4}{2^{10}.3^{10}.5^5}\)

= \(\dfrac{3^{10}.5^7.2^{10}}{2^{10}.3^{10}.5^5}\)

= 25
@trần phúc nguyên

A = \(\dfrac{72^3.5^{42}}{108^4}\)

= \(\dfrac{\left(2^3.3^2\right)^3.5^{42}}{\left(2^2.3^3\right)^4}\)

= \(\dfrac{2^9.3^6.5^{42}}{2^8.3^{12}}\)

= \(\dfrac{2.5^{42}}{3^6}\)

@trần phúc nguyên

a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)

= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)

= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)

= 1
@Nguyen Thi Ngoc Linh

c, Do (x - 3)(7 - x) > 0
=> x - 3; 7 - x cùng dấu
Xét 2 TH :
TH1 : \(\left\{{}\begin{matrix}x-3>0\\7-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\7>x\end{matrix}\right.\)<=> 3 < x < 7 (thỏa mãn)
TH2 : \(\left\{{}\begin{matrix}x-3< 0\\7-x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\7< x\end{matrix}\right.\)<=> 7 < x < 3 (vô lý)
Vậy x = 4; 5; 6
d, (x - 3)(x - 7) < 0
=> x - 3; x - 7 khác dấu
Xét 2 TH :
TH1 : \(\left\{{}\begin{matrix}x-3>0\\x-7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>3\\x< 7\end{matrix}\right.\)<=> 3 < x < 7 (thỏa mãn)
TH2 : \(\left\{{}\begin{matrix}x-3< 0\\x-7>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x>7\end{matrix}\right.\)<=> 7 < x < 3 (vô lý)
Vậy x = 4; 5; 6
@Jenny Jenny

Gọi số cần tìm có dạng là x = 21k + 6
Vì 21k \(⋮3\) và 6 \(⋮3\)
=> x \(⋮3;x>6\)
=> x là hợp số
@Nhok cute

a, Ta có : |1 - x| \(\ge0\Rightarrow\left|1-x\right|+8\ge8\)
Dấu ''='' xảy ra <=> 1 - x = 0 <=> x = 1
Vậy GTNN của A = 8 <=> x = 1
b, Do (x + 5)(x + 12) < 0
=> x + 5; x + 12 khác dấu
Xét 2 TH :
TH1 : \(\left\{{}\begin{matrix}x+5< 0\\x+12>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -5\\x>-12\end{matrix}\right.\)<=> -12 < x < -5 (thỏa mãn)
TH2 : \(\left\{{}\begin{matrix}x+5>0\\x+12< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-5\\x< -12\end{matrix}\right.\)<=> -5 < x < -12 (vô lý)
Vậy x = -11; -10; -9; -8; -7; -6
@Nhok cute

tại dai nhưng dài quá bạn l i k e rồi mình già hết ra cho

89 tick nháphùng ngọc ánh

Theo bài ra ta có :
a + b = 7(a - b)
=> a + b = 7a - 7b
=> b + 7b = 7a - a
=> 8b = 6a
=> b = \(\dfrac{6a}{8}=\dfrac{3a}{4}\) (1)
a.b = 192(a - b)
=> a.\(\dfrac{3a}{4}=192\left(a-\dfrac{3a}{4}\right)\)
=> \(\left\{{}\begin{matrix}a=0;64\\b=0;48\end{matrix}\right.\)
Vậy các cặp (a; b) thỏa mãn là (0; 0); (64; 48)
@Trần Thị Trà My