\(\dfrac{10.4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{5.2.2^{12}.3^{10}+3^9.2^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

\(=\dfrac{5.2^{13}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}=\dfrac{2^{12}.3^{10}.5\left(2+1\right)}{2^{11}.3^{11}.5}\)

\(=\dfrac{2^{11}.2.3^{10}.5.3}{2^{11}.3^{10}.3.5}=2\)

x = 3

x = 3

\(120-\left(3x-9\right)=1440:12\\ \Rightarrow120-\left(3x-9\right)=120\\ \Rightarrow3x-9=120-120\\ \Rightarrow3x-9=0\\ \Rightarrow3x=0+9\\ \Rightarrow3x=9\\ \Rightarrow x=9:3\\ \Rightarrow x=3\)

\(\frac{15}{x-9}=\frac{20}{y-12}=\frac{40}{z-24}\) và \(x.y=1200\) (Sửa đề)

Ta có:

\(\frac{x-9}{15}=\frac{y-12}{20}=\frac{z-24}{40}\Rightarrow\frac{x}{15}.\frac{9}{15}=\frac{y}{20}.\frac{12}{20}=\frac{z}{40}.\frac{24}{40}\)

Mà \(\frac{9}{15}=\frac{12}{20}=\frac{24}{40}=\frac{3}{5}\Rightarrow\frac{x}{15}=\frac{y}{20}=\frac{z}{40}\)

\(\Rightarrow\frac{x^2}{15^2}=\frac{y^2}{20^2}=\frac{z^2}{40^2}=\frac{x.y}{15.20}=\frac{1200}{300}=2^2\)

\(\Rightarrow x^2=2^2.15=\left(2.15\right)^2=30^2\Rightarrow x=\pm30\)

\(\Rightarrow y^2=2^2.20^2=\left(2.20\right)^2=40^2\Rightarrow y=\pm40\)

\(\Rightarrow z^2=2^2.40^2=\left(2.40\right)^2=80^2\Rightarrow z=\pm80\)

bạn muốn điểm hỏi đáp thì trả lời CÂU HỎI CỦA NGƯỜI KHÁC chứ ntn ít người tk lắm

ừ nhưng giờ tối rồi khó tìm hơn bạn à, để mai mình cố gắng

Câu 1: 

=>5x=15

hay x=3

Câu 2: 

\(\Leftrightarrow120-\left(3x+9\right)=12\)

=>3x+9=108

=>3x=99

hay x=33

\(A=\frac{4^6x9^5+6^9x120}{8^4x3^{12}-6^{11}}\)

\(A=\frac{2^{12}x3^{10}+2^9x3^9x2^3x3x5}{2^{12}x3^{12}-2^{11}x3^{11}}\)

\(A=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{10}-2^{11}.3^{11}}\)

\(A=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{10}.\left(2-3\right)}=\frac{2.6}{-1}=-12\)

cảm ơn bạn vì đã giúp mình

a,  3^9:3^2=3^7

b,  (3/5)^15:(3/5)^10=(3/5)^5

c,  (5.3.3)^10.5^10 / (5.5.3)^10=5^10.3^10.3^10.5^10 / 5^10.5^10.3^10=5^20.3^20 / 5^20.3^10=3^10

Đặt:

\(X=\left(1+\dfrac{1}{9}\right)\left(1+\dfrac{1}{10}\right)\left(1+\dfrac{1}{11}\right).....\left(1+\dfrac{1}{200}\right)\)

\(X=\dfrac{10}{9}.\dfrac{11}{10}.\dfrac{12}{11}......\dfrac{201}{200}\)

\(X=\dfrac{10.11.12......201}{9.10.11......200}\)

\(X=\dfrac{201}{9}\)

\(Y=\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{11}\right)\left(1-\dfrac{1}{12}\right).....\left(1-\dfrac{1}{99}\right)\)

\(Y=\dfrac{9}{10}.\dfrac{10}{11}.\dfrac{11}{12}.....\dfrac{98}{99}\)

\(Y=\dfrac{9.10.11......98}{10.11.12.....99}\)

\(Y=\dfrac{9}{99}=\dfrac{1}{11}\)

Giúp mình với mai mình phải nạp cho cô giáo