\(\dfrac{6^9.2^{10}+12^{10}}{2^{19}.27^3+15.4^9.9^4}\)

\(=\dfrac{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}\)

\(=\dfrac{2^{19}.3^9+2^{20}.3^{10}}{2^{19}.3^9+5.2^{18}.3^9}\)

\(=\dfrac{2^{19}.3^9\left(1+2.3\right)}{2^{19}.3^9\left(1+5.2\right)}\)

\(=\dfrac{1+6}{1+10}=\dfrac{7}{11}\)

\(\dfrac{6^9.2^{10}+12^{10}}{2^{19}.27^3+15.4^9.9^4}\\ =\dfrac{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9\left(3^2\right)^4}\\ =\dfrac{2^{19}.3^9+2^{20}.3^{10}}{2^{19}.3^9+5.2^{18}.3^9}\\ =\dfrac{2^{18}.3^9.\left(2+2^2.3\right)}{2^{18}.3^9.\left(2+5\right)}\\ =\dfrac{14}{7}\\ =2\)

- Muốn trừ hai phân số có cùng mẫu số, ta trừ tử số của phân số thứ nhất cho mẫu số của phân số thứ hai và giữ nguyên mẫu số.

- Muốn trừ hai phân số khác mẫu số, ta quy đồng mẫu số hai phân số, rồi trừ hai phân số đó.

\(\dfrac{-10}{9}-\dfrac{9}{8}\)

\(=-\dfrac{80}{72}-\dfrac{81}{72}\)

\(=-\dfrac{161}{72}\)

a) \(x+\dfrac{1}{3}=\dfrac{4}{7}\Leftrightarrow x=\dfrac{4}{7}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{21}\)

Vậy \(x=\dfrac{5}{21}\)

b) \(\dfrac{5}{9}-\dfrac{2}{3}x=\dfrac{1}{3}\Leftrightarrow\dfrac{5}{9}-\dfrac{1}{3}=\dfrac{2}{3}x\Leftrightarrow\dfrac{2}{3}x=\dfrac{2}{9}\Leftrightarrow x=\dfrac{2}{9}:\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3}\)

c) \(x-\dfrac{3}{4}=\dfrac{2}{7}\Leftrightarrow x=\dfrac{2}{7}+\dfrac{3}{4}\Leftrightarrow x=\dfrac{29}{28}\)

Vậy \(x=\dfrac{29}{28}\)

d) \(\left(\dfrac{5}{7}-x\right).\dfrac{11}{15}=-\dfrac{22}{45}\Leftrightarrow\dfrac{5}{7}-x=-\dfrac{22}{45}:\dfrac{11}{15}\Leftrightarrow\dfrac{5}{7}-x=-\dfrac{2}{3}\Leftrightarrow\dfrac{5}{7}+\dfrac{2}{3}=x\Leftrightarrow x=\dfrac{29}{21}\)Vậy \(x=\dfrac{29}{21}\)

a. \(x=\dfrac{4}{7}-\dfrac{1}{3}\)

\(x=\dfrac{12}{21}-\dfrac{7}{21}\)

\(x=\dfrac{5}{21}\)

Vậy \(x=\dfrac{5}{21}\)

b. \(-\dfrac{2}{3}x=\dfrac{1}{3}-\dfrac{5}{9}\)

\(-\dfrac{2}{3}x=\dfrac{3}{9}-\dfrac{5}{9}\)

\(-\dfrac{2}{3}x=\dfrac{-2}{9}\)

\(x=\dfrac{-2}{9}:\left(\dfrac{-2}{3}\right)\)

\(x=\dfrac{-2}{9}.\dfrac{-3}{2}\)

\(x=\dfrac{1}{3}\)

\(vậyx=\dfrac{1}{3}\)

c. \(x=\dfrac{2}{7}+\dfrac{3}{4}\)

\(x=\dfrac{8}{28}+\dfrac{21}{28}\)

\(x=\dfrac{29}{28}\)

\(vẫy=\dfrac{29}{28}\)

d. \(\left(\dfrac{5}{7}-x\right).\dfrac{11}{15}=\dfrac{-22}{45}\)

\(\left(\dfrac{5}{7}-x\right)=\dfrac{-22}{45}.\dfrac{15}{11}\)

\(\left(\dfrac{5}{7}-x\right)=\dfrac{-2}{3}\)

\(-x=\dfrac{2}{3}+\dfrac{5}{7}\)

\(-x=\dfrac{-14}{21}+\dfrac{15}{21}\)

\(-x=\dfrac{-29}{21}\)

\(x=\dfrac{29}{21}\)

\(vẫy=\dfrac{29}{21}\)

mk ko ghi lại đề nha

\(\dfrac{3^{18}.24^4}{18^4.9^5}=\dfrac{3^{18}.6^4.4^4}{3^4.6^4.\left(3^2\right)^5}\\ =\dfrac{3^{18}.6^4.4^4}{3^4.6^4.3^{10}}\\ =\dfrac{3^{18}.6^4.4^4}{3^{14}.6^4}=3^4.4^4=12^4\)

\(\dfrac{3^{18}.24^4}{18^4.9^5}=\dfrac{3^{18}.3^4.8^4}{9^4.2^4.9^5}=\dfrac{3^{22}.\left(2^3\right)^4}{\left(3^2\right)^9.2^4}=\dfrac{3^{22}.2^{12}}{3^{18}.2^4}=3^4.2^8\)

\(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x\left(1+3+3^2\right)=117\)

\(3^x=\dfrac{117}{13}=9=3^2\)

\(\Rightarrow x=2\)

Cám ơn bạn nha

Ta có:

\(3^x+3^{x+1}+3^{x+2}=117\)

\(\Leftrightarrow3^x+3^x.3+3^x.3^2=117\)

\(\Leftrightarrow3^x\left(1+3+3^2\right)=117\)

\(\Leftrightarrow3^x.13=117\Leftrightarrow3^x=9\Leftrightarrow3^x=3^2\)

\(\Leftrightarrow x=2\)

Vậy...

Cám ơn nha

Ta có \(x^2\ge0;\left|x-1\right|\ge0\)

\(\Rightarrow x^2+\left|x-1\right|\ge0\)

\(\Rightarrow\) |x² + |x-1|| = x²+|x-1|

\(\Rightarrow x^2+\left|x-1\right|=x^2+2\Leftrightarrow\left|x-1\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Cám ơn bạn

\(\Delta ABC=\Delta HIK\)

\(\Rightarrow\left\{{}\begin{matrix}AB=HI;BC=IK;AC=HK\\\widehat{A}=\widehat{H};\widehat{B}=\widehat{I};\widehat{C}=\widehat{K}\end{matrix}\right.\)

như vậy ta có thể suy ra được:

\(HI=2cm;IK=40cm;\widehat{I}=40^0\)