- 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:

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

A C B H I K

a, Cạnh tương ứng với canh BC là cạnh IK

Góc tương ứng với góc H là góc : góc A

b, các cạnh bằng nhau là: AB = HI; AC = HK ; BC = IK

các cặp góc bằng nhau là: góc A = góc H; góc B = góc I; góc C = góc K