『Kuroba ム Tsuki Ryoo ︵²ᵏ⁷』
Giới thiệu về bản thân
\(\text{∘ Ans}\)
\(\downarrow\)
`1,`
`a)`
`15,27 - 4,18 - 2,09`
`= (15,27 - 2,09) - 4,18`
`= 13,18 - 4,18`
`= 9`
`b)`
`60 - 26,75 - 13,25`
`= 60 - (26,75 + 13,25)`
`= 60 - 40 = 20`
`c)`
`38,25 - 18,25 + 21,64 - 11,64 + 9,93`
`= (38,25 - 18,25) + (21,64 - 11,64) + 9,93`
`= 20 + 10 + 9,93`
`= 30 - 9,93`
`=``20,07`
`e)`
`(72,69 + 18,47 ) - (8,47 + 22,69)`
`= 72,69 + 18,47 - 8,47 - 22,69`
`= (72,69 - 22,69) + (18,47 - 8,47)`
`= 50 + 10 = 60`
`2,`
Chiều dài của HCN đó là:
`36 \div \dfrac{3}{5} = 60 (cm)`
Đổi `60 cm = 0,6 m`; `36 cm = 0,36 m`
Sợi dây thép dài số m là:
`(0,6 + 0,36) \times 2 = 1,92 (m)`
Đáp số: `1,92m.`
\(\text{∘ Ans}\)
\(\downarrow\)
`(x - 2).(x - 3) = 0`
`\Rightarrow`
`\text {TH1: } x - 2 = 0`
`\Rightarrow x = 0 + 2`
`\Rightarrow x = 2`
`\text {TH2: } x - 3 = 0`
`\Rightarrow x = 0 + 3`
`\Rightarrow x = 3`
Vậy, `x \in`\(\left\{2;3\right\}\)
\(\text{∘ Ans}\)
`A =`\(\left(\dfrac{27}{23}+\dfrac{1}{2}\right)-\left(\dfrac{5}{21}+\dfrac{4}{23}\right)-\dfrac{37}{21}\)
`=`\(\dfrac{27}{23}+\dfrac{1}{2}-\dfrac{5}{21}-\dfrac{4}{23}-\dfrac{37}{21}\)
`=`\(\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(-\dfrac{5}{21}-\dfrac{37}{21}\right)+\dfrac{1}{2}\)
`=`\(\dfrac{23}{23}-\dfrac{42}{21}+\dfrac{1}{2}=1-2+0,5=1-1,5=-0,5\)
\(\text{∘ Ans}\)
\(\downarrow\)
`10^2 - [ 60 \div (5^6 \div 5^4 - 3.5)]`
`= 10^2 - [60 \div (5^2 - 3.5)]`
`= 10^2 - {60 \div [5(5 - 3)]}`
`= 10^2 - [60 \div (5.2)]`
`= 10^2 - (60 \div 10)`
`= 10^2 - 6`
`= 100 - 6`
`= 94`
\(\text{∘ Ans}\)
\(\downarrow\)
Bạn đăng CH đúng môn học nha!
`4,`
\(3\dfrac{1}{5}=\dfrac{5\times3+1}{5}=\dfrac{16}{5}\)
\(\Rightarrow C\)
`5,`
Mười sáu đơn vị: `16`
Bốn trăm chín mươi hai phần nghìn: `492`
\(\Rightarrow\) `16,492`
\(\Rightarrow\) `B`
`6,`
\(\dfrac{40}{7}\cdot\dfrac{14}{5}=\dfrac{5\times8}{7}\cdot\dfrac{7\times2}{5}=\dfrac{8}{1}\cdot\dfrac{2}{1}=\dfrac{16}{1}=16\)
\(\Rightarrow\) \(C\)
`7,`
\(2\dfrac{1}{4}=\dfrac{2\times4+1}{4}=\dfrac{9}{4}\)
\(\dfrac{9}{4}\div\dfrac{1}{8}=18\)
Vậy, \(2\dfrac{1}{4}\) gấp `18` lần \(\dfrac{1}{8}\)
\(\Rightarrow B\)
`8,`
Chiều rộng hình chữ nhật là:
\(\dfrac{25}{12}\div\dfrac{5}{3}=\dfrac{5}{4}\left(m\right)\)
Chu vi hình chữ nhật đó là:
\(\left(\dfrac{5}{3}+\dfrac{5}{4}\right)\times2=\dfrac{35}{6}\left(m\right)\)
Ta có: \(\dfrac{35}{6}=\dfrac{\left(35-5\right)\div6}{6}=5\dfrac{5}{6}\)
\(\Rightarrow B.\)
\(\text{∘ Ans}\)
\(\downarrow\)
\(A=\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
`=`\(\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\left[1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}\right]\)
`=`\(\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
`=`\(\dfrac{8}{9}-\dfrac{8}{9}=0\)
Vậy, ` A = 0.`