Câu hỏi của Nguyễn Quang Tùng

Question

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『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』 · Accepted Answer

`#3107.101107`

`1.`

`a)`

$\dfrac{7}{23}-\dfrac{5}{14}+\dfrac{1}{2}-\dfrac{9}{14}+\dfrac{16}{23}\ =\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\left(\dfrac{5}{14}+\dfrac{9}{14}\right)+\dfrac{1}{2}\ =\dfrac{23}{23}-\dfrac{14}{14}+\dfrac{1}{2}\ =1-1+\dfrac{1}{2}\ =\dfrac{1}{2}$

`b)`

$\left(-\dfrac{2}{3}\right)\cdot\dfrac{3}{11}+\left(-\dfrac{16}{9}\right)\cdot\dfrac{3}{11}\ =\dfrac{3}{11}\cdot\left(-\dfrac{2}{3}-\dfrac{16}{9}\right)\ =\dfrac{3}{11}\cdot\left(-\dfrac{22}{9}\right)\ =-\dfrac{2}{3}$

`2.`

`a)`

$-\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{1}{3}\ \Rightarrow-\dfrac{3}{7}x=\dfrac{1}{6}\ \Rightarrow x=\dfrac{1}{6}\div\left(-\dfrac{3}{7}\right)\ \Rightarrow x=-\dfrac{7}{18}$

`b)`

$\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\x=-\dfrac{3}{4}\end{matrix}\right.$

_____

`1.`

`a)`

$-\dfrac{9}{17}+6,72+\dfrac{-8}{17}+\left(-4,72\right)\ =\left(-\dfrac{9}{17}-\dfrac{8}{17}\right)+\left(6,72-4,72\right)\ =-\dfrac{17}{17}+2\ =-1+2=1$

`b)`

$2\dfrac{1}{5}-\left(-\dfrac{3}{4}+\dfrac{1}{5}\right)\ =\dfrac{11}{5}+\dfrac{3}{4}-\dfrac{1}{5}\ =\left(\dfrac{11}{5}-\dfrac{1}{5}\right)+\dfrac{3}{4}\ =\dfrac{10}{5}-\dfrac{3}{4}\ =2-\dfrac{3}{4}\ =\dfrac{5}{4}$

`c)`

$\left(-\dfrac{1}{3}\right)^3-\dfrac{3}{8}\div \left(\dfrac{1}{2}\right)^3-\dfrac{5}{2}\cdot\left(-2\right)\ =\left(-\dfrac{1}{27}\right)-\dfrac{3}{8}\div\dfrac{1}{8}-\left(-5\right)\ =-\dfrac{1}{27}-3+5\ =-\dfrac{1}{27}+2\ =\dfrac{53}{27}$

`d)`

$4,1\cdot\dfrac{-5}{12}-6,2+4,1\cdot\dfrac{-7}{12}\ =4,1\cdot\left(-\dfrac{5}{12}-\dfrac{7}{12}\right)-6,2\ =4,1\cdot\left(-\dfrac{12}{12}\right)-6,2\ 4,1\cdot\left(-1\right)-6,2\ =-4,1-6,2\ =-10,3$

Câu trả lời:

`#3107.101107`

`1.`

`a)`

$\dfrac{7}{23}-\dfrac{5}{14}+\dfrac{1}{2}-\dfrac{9}{14}+\dfrac{16}{23}\ =\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\left(\dfrac{5}{14}+\dfrac{9}{14}\right)+\dfrac{1}{2}\ =\dfrac{23}{23}-\dfrac{14}{14}+\dfrac{1}{2}\ =1-1+\dfrac{1}{2}\ =\dfrac{1}{2}$

`b)`

$\left(-\dfrac{2}{3}\right)\cdot\dfrac{3}{11}+\left(-\dfrac{16}{9}\right)\cdot\dfrac{3}{11}\ =\dfrac{3}{11}\cdot\left(-\dfrac{2}{3}-\dfrac{16}{9}\right)\ =\dfrac{3}{11}\cdot\left(-\dfrac{22}{9}\right)\ =-\dfrac{2}{3}$

`2.`

`a)`

$-\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{1}{3}\ \Rightarrow-\dfrac{3}{7}x=\dfrac{1}{6}\ \Rightarrow x=\dfrac{1}{6}\div\left(-\dfrac{3}{7}\right)\ \Rightarrow x=-\dfrac{7}{18}$

`b)`

$\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{4}\right)^2\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}-\dfrac{1}{2}\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{matrix}\right.\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\x=-\dfrac{3}{4}\end{matrix}\right.$

_____

`1.`

`a)`

$-\dfrac{9}{17}+6,72+\dfrac{-8}{17}+\left(-4,72\right)\ =\left(-\dfrac{9}{17}-\dfrac{8}{17}\right)+\left(6,72-4,72\right)\ =-\dfrac{17}{17}+2\ =-1+2=1$

`b)`

$2\dfrac{1}{5}-\left(-\dfrac{3}{4}+\dfrac{1}{5}\right)\ =\dfrac{11}{5}+\dfrac{3}{4}-\dfrac{1}{5}\ =\left(\dfrac{11}{5}-\dfrac{1}{5}\right)+\dfrac{3}{4}\ =\dfrac{10}{5}-\dfrac{3}{4}\ =2-\dfrac{3}{4}\ =\dfrac{5}{4}$

`c)`

$\left(-\dfrac{1}{3}\right)^3-\dfrac{3}{8}\div \left(\dfrac{1}{2}\right)^3-\dfrac{5}{2}\cdot\left(-2\right)\ =\left(-\dfrac{1}{27}\right)-\dfrac{3}{8}\div\dfrac{1}{8}-\left(-5\right)\ =-\dfrac{1}{27}-3+5\ =-\dfrac{1}{27}+2\ =\dfrac{53}{27}$

`d)`

$4,1\cdot\dfrac{-5}{12}-6,2+4,1\cdot\dfrac{-7}{12}\ =4,1\cdot\left(-\dfrac{5}{12}-\dfrac{7}{12}\right)-6,2\ =4,1\cdot\left(-\dfrac{12}{12}\right)-6,2\ 4,1\cdot\left(-1\right)-6,2\ =-4,1-6,2\ =-10,3$

『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』 · Answer

`2.`

`a)`

$x+\left(-\dfrac{1}{9}\right)=-\dfrac{7}{6}\ \Rightarrow x=-\dfrac{7}{6}-\left(-\dfrac{1}{9}\right)\ \Rightarrow x=-\dfrac{7}{6}+\dfrac{1}{9}\ \Rightarrow x=-\dfrac{19}{18}$

`b)`

$\left(x-\dfrac{4}{7}\right)\div\dfrac{-5}{3}=0,2\ \Rightarrow x-\dfrac{4}{7}=0,2\cdot\left(-\dfrac{5}{3}\right)\ \Rightarrow x-\dfrac{4}{7}=-\dfrac{1}{3}\ \Rightarrow x=-\dfrac{1}{3}+\dfrac{4}{7}\ \Rightarrow x=\dfrac{5}{21}$

`c)`

$\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{8}\ \Rightarrow\left(2x-\dfrac{1}{3}\right)^3=\left(\dfrac{1}{2}\right)^3\ \Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\ \Rightarrow2x=\dfrac{1}{2}+\dfrac{1}{3}\ \Rightarrow2x=\dfrac{5}{6}\ \Rightarrow x=\dfrac{5}{6}\div2\ \Rightarrow x=\dfrac{5}{12}$

____

`13.`

`a)`

Ta có: $\dfrac{139}{303}>\dfrac{138}{303}=\dfrac{46}{101}$

Mà $\dfrac{35}{101}< \dfrac{46}{101}$

$\Rightarrow\dfrac{139}{303}>\dfrac{35}{101}$

`b)`

Ta có:

$\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{8-2}=\left(\dfrac{1}{2}\right)^6$

$\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3=\left(\dfrac{1}{2}\right)^{3+3}=\left(\dfrac{1}{2}\right)^6$

Vì $\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6\ \Rightarrow\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3$

`14.`

`a)`

$-\dfrac{11}{24}+\dfrac{3}{4}-\dfrac{13}{24}\ =\left(-\dfrac{11}{24}-\dfrac{13}{24}\right)+\dfrac{3}{4}\ =-\dfrac{24}{24}+\dfrac{3}{4}\ =-1+\dfrac{3}{4}\ =-\dfrac{1}{4}$

`b)`

$-\dfrac{5}{9}-\left(\dfrac{8}{15}+\dfrac{4}{9}\right)+\dfrac{7}{15}\ =-\dfrac{5}{9}-\dfrac{8}{15}-\dfrac{4}{9}+\dfrac{7}{15}\ =\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)-\left(\dfrac{8}{15}+\dfrac{7}{15}\right)\ =-\dfrac{9}{9}-\dfrac{15}{15}\ =-1-1=-2$

`c)`

$\dfrac{5}{9}\div2,4-\dfrac{41}{9}\div2,4\ =\dfrac{5}{9}\div\dfrac{12}{5}-\dfrac{41}{9}\div\dfrac{12}{5}\ =\dfrac{5}{9}\cdot\dfrac{5}{12}-\dfrac{41}{9}\cdot\dfrac{5}{12}\ =\dfrac{5}{12}\cdot\left(\dfrac{5}{9}-\dfrac{41}{9}\right)\ =\dfrac{5}{12}\cdot\left(-\dfrac{36}{9}\right)\ =\dfrac{5}{12}\cdot\left(-4\right)\ =-\dfrac{5}{3}$

`a)`

`b)`

`c)`

Câu trả lời:

`2.`

`a)`

$x+\left(-\dfrac{1}{9}\right)=-\dfrac{7}{6}\ \Rightarrow x=-\dfrac{7}{6}-\left(-\dfrac{1}{9}\right)\ \Rightarrow x=-\dfrac{7}{6}+\dfrac{1}{9}\ \Rightarrow x=-\dfrac{19}{18}$

`b)`

$\left(x-\dfrac{4}{7}\right)\div\dfrac{-5}{3}=0,2\ \Rightarrow x-\dfrac{4}{7}=0,2\cdot\left(-\dfrac{5}{3}\right)\ \Rightarrow x-\dfrac{4}{7}=-\dfrac{1}{3}\ \Rightarrow x=-\dfrac{1}{3}+\dfrac{4}{7}\ \Rightarrow x=\dfrac{5}{21}$

`c)`

$\left(2x-\dfrac{1}{3}\right)^3=\dfrac{1}{8}\ \Rightarrow\left(2x-\dfrac{1}{3}\right)^3=\left(\dfrac{1}{2}\right)^3\ \Rightarrow2x-\dfrac{1}{3}=\dfrac{1}{2}\ \Rightarrow2x=\dfrac{1}{2}+\dfrac{1}{3}\ \Rightarrow2x=\dfrac{5}{6}\ \Rightarrow x=\dfrac{5}{6}\div2\ \Rightarrow x=\dfrac{5}{12}$

____

`13.`

`a)`

Ta có: $\dfrac{139}{303}>\dfrac{138}{303}=\dfrac{46}{101}$

Mà $\dfrac{35}{101}< \dfrac{46}{101}$

$\Rightarrow\dfrac{139}{303}>\dfrac{35}{101}$

`b)`

Ta có:

$\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^{8-2}=\left(\dfrac{1}{2}\right)^6$

$\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3=\left(\dfrac{1}{2}\right)^{3+3}=\left(\dfrac{1}{2}\right)^6$

Vì $\left(\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^6\ \Rightarrow\left(\dfrac{1}{2}\right)^8\div\left(\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^3\cdot\left(\dfrac{1}{2}\right)^3$

`14.`

`a)`

$-\dfrac{11}{24}+\dfrac{3}{4}-\dfrac{13}{24}\ =\left(-\dfrac{11}{24}-\dfrac{13}{24}\right)+\dfrac{3}{4}\ =-\dfrac{24}{24}+\dfrac{3}{4}\ =-1+\dfrac{3}{4}\ =-\dfrac{1}{4}$

`b)`

$-\dfrac{5}{9}-\left(\dfrac{8}{15}+\dfrac{4}{9}\right)+\dfrac{7}{15}\ =-\dfrac{5}{9}-\dfrac{8}{15}-\dfrac{4}{9}+\dfrac{7}{15}\ =\left(-\dfrac{5}{9}-\dfrac{4}{9}\right)-\left(\dfrac{8}{15}+\dfrac{7}{15}\right)\ =-\dfrac{9}{9}-\dfrac{15}{15}\ =-1-1=-2$

`c)`

$\dfrac{5}{9}\div2,4-\dfrac{41}{9}\div2,4\ =\dfrac{5}{9}\div\dfrac{12}{5}-\dfrac{41}{9}\div\dfrac{12}{5}\ =\dfrac{5}{9}\cdot\dfrac{5}{12}-\dfrac{41}{9}\cdot\dfrac{5}{12}\ =\dfrac{5}{12}\cdot\left(\dfrac{5}{9}-\dfrac{41}{9}\right)\ =\dfrac{5}{12}\cdot\left(-\dfrac{36}{9}\right)\ =\dfrac{5}{12}\cdot\left(-4\right)\ =-\dfrac{5}{3}$

`a)`

`b)`

`c)`

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