Tính ra thừa số nguyên tố

\(A=\frac{2^9.3^6:2^2:3^6}{2^2.271}\)

\(A=\frac{2^5}{271}=\frac{32}{271}\)

1084 +(942 -2x) :36 x44 =1434

=>(942 -2x) :36 x44 =1434 -1084

=>(942 -2x) :36 x44 = 350

=>(942 -2x) :36 =350 :44

=>(942 -2x) :36 =175 /22

=>(942 -2x)=175 /22x 36

=>(942 -2x)= 3150 /11

=>2x = 3150 /11-942

=>2x =-7212 /11

kết quả nè còn lại bn tự nghĩ nhé! -655,6363636

xin lôi nhưng mink bn rui k ghi dc

\(1084+\left(942-2x\right)\div36\cdot44=1434\)

\(\left(942-2x\right)\div36\cdot44=1434-1084\)

\(\left(942-2x\right)\div36\cdot44=350\)

\(\left(942-2x\right)\div36=350\div44\)

\(\left(942-2x\right)\div36=\frac{175}{22}\)

\(942-2x=\frac{175}{22}\cdot36\)

\(942-2x=\frac{3150}{11}\)

\(2x=942-\frac{3150}{11}\)

\(2x=\frac{7212}{11}\)

\(\Rightarrow x=\frac{3606}{11}\)

\(C=\frac{72^3.54^2}{1084}\)

 \(=\frac{\left(8.9\right)^3.\left(2.27\right)^2}{1084}\)

\(=\frac{8^3.9^3.2^2.27^2}{1084}\)

\(=\frac{\left(2^3\right)^3.\left(3^2\right)^3.2^2.\left(3^3\right)^2}{1084}\)

\(=\frac{2^9.3^6.2^2.3^6}{1084}\)

\(=\frac{2^{11}.3^{12}}{1084}\)

\(=\frac{2^9.4.3^{12}}{271.4}\)

\(=\frac{2^9.3^{12}}{271}\)

bang 1004050.893

ket qua xau qua

nho h cho mik nha

1004050,893

các bạn hãy chỉ cách làm giúp mình

=373248.2916:1084

=1088391168:1084

=1004050,893