đề =\(\frac{\left(3^2\right)^{14}.\left(15^2\right)^5.\left(2^3\right)^7}{\left(2.9\right)^{12}.\left(5^3\right)^3.\left(8.3\right)^3}=\frac{3^{28}.15^{10}.2^{21}}{2^{12}.9^{12}.5^9.\left(2^3\right)^3.3^3}\) 

\(=\frac{3^{28}.3^{10}.5^{10}.2^{21}}{2^{21}.3^{27}.5^9}=\frac{3^{38}.5^{10}.2^{21}}{2^{21}.3^{27}.5^9}=3^9.5\) 

cái này bạn bấm mt là ra nha

k cho mình nha bạn

Đề: \(\frac{9^{14}\cdot225^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}\)

\(\Rightarrow\frac{\left(3^2\right)^{14}\cdot\left(5.3\right)^5\cdot\left(2^3\right)^7}{\left(3^2\cdot2\right)^{12}\cdot\left(5^4\right)^3\cdot\left(2^3\cdot3\right)^3}\)

\(\Rightarrow\frac{3^{28}\cdot5^5\cdot3^5\cdot2^{21}}{3^{24}\cdot2^{12}\cdot5^{12}\cdot2^9\cdot3^3}\)

\(\Rightarrow\frac{3^{33}\cdot5^5\cdot2^{21}}{3^{27}\cdot2^{21}\cdot5^{12}}\)

\(\Rightarrow\frac{3^{26}}{5^7}\)

Đề : \(\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)

\(\Rightarrow\frac{\left(3\cdot7\cdot13\cdot37\right)\cdot39-10101}{505050-70707}\)

\(\Rightarrow\frac{10101\cdot39-10101}{505050-70707}\)

\(\Rightarrow\frac{10101\cdot\left(39-1\right)}{505050-70707}\)

\(\Rightarrow\frac{10101\cdot38}{50\cdot10101-7\cdot10101}\)

\(\Rightarrow\frac{10101\cdot38}{10101\cdot\left(50-7\right)}\)

\(\Rightarrow\frac{10101\cdot38}{10101\cdot43}\)

\(\Rightarrow\frac{38}{43}\)

\(a,\left(x\cdot6\frac{2}{7}+\frac{3}{7}\right)\cdot2\frac{1}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(x\cdot\frac{44}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\left(\frac{44x}{7}+\frac{3}{7}\right)\cdot\frac{11}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\frac{44x+3}{7}\cdot\frac{11}{5}-\frac{3}{7}=-2\)

\(\Rightarrow\frac{11\cdot\left(44x+3\right)}{5\cdot7}-\frac{3}{7}=-2\)

\(\Rightarrow\frac{484x+33}{35}-\frac{3}{7}=-2\)

\(\Rightarrow\frac{484x+33}{35}-\frac{15}{35}=-2\)

\(\Rightarrow\frac{484x+33-15}{35}=-2\)

\(\Rightarrow\frac{484x+18}{35}=-2\)

\(\Rightarrow\frac{484x+18}{35}=\frac{-70}{35}\)

\(\Rightarrow484x+18=\left(-70\right)\)

\(\Rightarrow484x=\left(-70\right)-18\)

\(\Rightarrow484x=-88\)

\(\Rightarrow x=-\frac{88}{484}=-\frac{2}{11}\)

\(b,x\cdot3\frac{1}{4}+\left(-\frac{7}{6}\right)\cdot x-1\frac{2}{3}=\frac{5}{12}\)

\(\Rightarrow x\cdot\frac{13}{4}+\left(-\frac{7}{6}\right)\cdot x-\frac{5}{3}=\frac{5}{12}\)

\(\Rightarrow\frac{13x}{4}+\left(-\frac{7x}{6}\right)-\frac{5}{3}=\frac{5}{12}\)

\(\Rightarrow\frac{13x\cdot3}{12}+\left(-\frac{7x\cdot2}{12}\right)-\frac{5\cdot4}{12}=\frac{5}{12}\)

\(\Rightarrow\frac{39x}{12}+\left(-\frac{14x}{12}\right)-\frac{20}{12}=\frac{5}{12}\)

\(\Rightarrow\frac{39x-14x-20}{12}=\frac{5}{12}\)

\(\Rightarrow\frac{25x-20}{12}=\frac{5}{12}\)

\(\Rightarrow25x-20=5\)

\(\Rightarrow25x=20+5\)

\(\Rightarrow25x=25\)

\(\Rightarrow x=1\)

ĐKXĐ : x khác 1

ta có \(A=\frac{3x}{1-x}=\frac{-3x}{x-1}=\frac{-3x-3+3}{x-1}=\frac{-3\left(x-1\right)+3}{x-1}\) 

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

vậy để A thuộc Z thì x-1 thuộc Ư(3) => x-1 thuộc {-3;-1;1;3} => x thuộc {-2;0;2;4) vì tất cả đều thỏa mãn ĐKXĐ

vậy để A thuộc Z thì x phải có các giá trị là : -2;0;2;4

bạn k cho mình nha

Chiều rộng của khu vườn là: 

120*1/2=60(m)

Chu vi khu vườn là:

(120+60)*2 = 320(m2)

                  =0,032(ha)

\(a)\)Chiều rộng khu vườn đó là:

\(120\times\frac{1}{2}=60\left(m\right)\)

Chu vi khu vườn đó là:

\(\left(120+60\right)\times2=360\left(m\right)\)

\(b)\)Diện tích khu vườn đó là:

\(120\times60=7200\left(m^2\right)\)

\(7200m^2=0,72ha\)

\(Đ/s:.....\)

x+5,22=9,08

x=9.08-5.22

x=3.86

x = 9,08 - 5,22 
x = 3,86

17,54+x:3,5=64,16

x:3,5=64,16-17.54

x:3,5=46.62

x=46.62:3.5

x=13.32

(x-1,75):0,01:20=41,25

(x-1,75):0,01=41.25x20

(x-1,75):0,01=825

x-1.75=825x0.01

x-1.75=82.5

x=82.5-1.75

x=80.75

A/

 \(17,54+\frac{x}{3,5}=64,16\)

\(\Rightarrow\frac{x}{3,5}=64,16-17,54\)

\(\Rightarrow\frac{x}{3,5}=46,62\)

\(\Rightarrow x=46,62\cdot3,5\)

\(\Rightarrow x=163,17\)

B/ \(\left(x-1,75\right):0,01:20=41,25\)

\(\Rightarrow\left(x-1,75\right):0,01=41,25\cdot20\)

\(\Rightarrow\left(x-1,75\right):0,01=825\)

\(\Rightarrow\left(x-1,75\right)=825\cdot0,01\)

\(\Rightarrow x-1,75=8,25\)

\(\Rightarrow x=8,25+1,75\)

\(\Rightarrow x=10\)

Add: Tr Ph Thảo (hpthaoo)

Gọi AF giao BC tại G. Theo ĐL Thales thì \(\frac{FA}{FG}=\frac{ED}{EB}=1\), suy ra F là trung điểm AG

Dễ thấy tam giác ABG cân tại B,do đó AG vuông góc BF

Đường thẳng AG: đi qua \(F\left(4;3\right)\), VTPT \(\overrightarrow{FB}=\left(1;-2\right)\)\(\Rightarrow AG:x-2y+2=0\)

Xét hệ \(\hept{\begin{cases}x+2y-18=0\\x-2y+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\y=5\end{cases}}\Rightarrow A\left(8;5\right)}\)

Vì F là trung điểm AG nên \(G\left(0;1\right)\)\(\Rightarrow\overrightarrow{GB}=\left(5;0\right)\)=> VTPT của BC là \(\left(0;1\right)\)

\(\Rightarrow BC:x-1=0\). Vậy \(d\left(O;BC\right)=1.\)