4⁵.3² + 2¹⁰ - 2³⁰

= (2²)⁵.9 + 2¹⁰ - 2³⁰

= 2¹⁰.9 + 2¹⁰ - 2³⁰

= 2¹⁰.(9 + 1 - 2²⁰)

= 2¹⁰.(10 - 1048576)

= 2¹⁰.(-1048566)

= -1073731548

Ko bít

Mày là

Ta có:

\(2x=3y=5z\)\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{3y}{30}=\dfrac{2z}{12}=\dfrac{x+3y-2z}{15+30-12}=\dfrac{66}{33}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=30\\y=20\\z=12\end{matrix}\right.\)

 Đổi với chương trình lớp 7 thì chị nên thêm câu "Áp dụng tính chất dãy tỉ số bằng nhau ta có: " nhé 

\(1,8=\dfrac{18}{10}=\dfrac{9}{5}\)

\(1,8=\dfrac{18}{10}=\dfrac{18:2}{10:2}=\dfrac{9}{5}\)

\(\dfrac{7}{2\cdot9}+\dfrac{7}{9\cdot16}+....+\dfrac{7}{86\cdot93}=\dfrac{\overline{a1}}{\overline{bcd}}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{86}-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)

\(\Rightarrow\dfrac{1}{2}-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-\left(\dfrac{1}{16}-\dfrac{1}{16}\right)-...-\left(\dfrac{1}{86}-\dfrac{1}{86}\right)-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{93}=\dfrac{\overline{a1}}{\overline{bcd}}\)

\(\Rightarrow\dfrac{91}{186}=\dfrac{\overline{a1}}{\overline{bcd}}\)

(1): \(\overline{a1}=91\Rightarrow a=9\)

(2): \(\overline{bcd}=186\Rightarrow\left\{{}\begin{matrix}b=1\\c=8\\d=6\end{matrix}\right.\)

Vậy: ... 

Cách 1: 

\(6:2\cdot\left(1+2\right)\)

\(=\left(6:2\right)\cdot\left(1+2\right)\)

\(=3\cdot3\)

\(=9\)

Cách 2:

\(6:2\cdot\left(1+2\right)\)

\(=6:\left[2:\left(1+2\right)\right]\)

\(=6:\left(2:3\right)\)

\(=6:\dfrac{2}{3}\)

\(=6\cdot\dfrac{3}{2}\)

\(=3\cdot3\)

\(=9\)

Vậy kết quả đúng của phép tính là bằng 9 

1

\(4\left(2x+1\right)^2=576\)

\(\left(2x+1\right)^2=\dfrac{576}{4}=144=12^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=11\\2x=-13\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\)

\(4\cdot(2x+1)^2=576\\\Rightarrow (2x+1)^2=576:4\\\Rightarrow(2x+1)^2=144\\\Rightarrow(2x+1)^2=(\pm12)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=12\\2x+1=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=11\\2x=-13\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=-\dfrac{13}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\dfrac{13}{2};\dfrac{11}{2}\right\}\)

\(\dfrac{3}{31}+\left(\dfrac{1}{6}+\dfrac{-5}{9}\right):\left(\dfrac{7}{24}-\dfrac{13}{18}\right)\)

\(=\dfrac{3}{31}+\dfrac{3-5.2}{18}:\dfrac{7.3-13.4}{72}\)

\(=\dfrac{3}{31}+\dfrac{-7}{18}:\dfrac{-31}{72}\)

\(=\dfrac{3}{31}+\dfrac{7.72}{18.31}=\dfrac{3}{31}+\dfrac{28}{31}=1\)