`#3107`

`1,`

`a)` Yc là tính \(\widehat{yOm}\) và \(\widehat{xOn}\) phải k c? Ba điểm x, O, m cùng nằm trên 1 đt' mà?

Vì \(\widehat{xOy}\) và \(\widehat{xOn}\) là 2 góc kề bù

\(\widehat{xOy} + \widehat{xOn} = 180^0 \\ \Rightarrow 50^0 + \widehat{xOn} = 180^0 \\ \Rightarrow \widehat{xOn} = 130^0\)

Vì \(\widehat{xOn}\) và \(\widehat{yOm}\) là 2 góc đối đỉnh

`=>`$ \widehat{xOn} = \widehat{yOm} = 130^0$

Vậy, $ \widehat{xOn} = \widehat{yOm} = 130^0.$

`b)`

Vì Oz là tia phân giác của $\widehat{yOm}$

`=>` $\widehat{mOz} = \widehat{yOz} = \dfrac{1}{2} \widehat{yOm}$

`=>` $\widehat{mOz} = \dfrac{1}{2} . 130^0 = 65^0$

Vậy, $\widehat{mOz} = 65^0.$

Diện tích xung quanh bể cá:

\(25\cdot25\cdot4=2500\left(dm^2\right)=\left(25m^2\right)\)

Thể tích bể cá: 

\(25\cdot25\cdot25=15625\left(dm^2\right)\)

Diện tích xung quanh của bể cá là:

\(25\times25\times4=2500\left(dm^2\right)\)

Thể tích của bể cá đó là:

\(25\times25\times25=15625\left(dm^3\right)\)

Tính giá trị của biểu thức sau ( hợp lí nếu có thể ):

a) \(\left(-25\right)\times\left(-5\right)\times\left(-0,4\right)\times\left(-0,2\right)\)

\(=\left[25\times\left(-0,2\right)\right]\times\left[5\times\left(-0,4\right)\right]\)

\(=\left(-5\right)\times2\)

\(=-10\)

b) \(\left(-0,25\right)\times0,02\times40\times\left(-50\right)\times\left(-201,43\right)\)

\(=\left(-0,25\times40\right)\times\left(-50\times0,02\right)\times\left(-201,43\right)\)

\(=\left(-10\right)\times\left(-1\right)\times\left(-201,43\right)\)

\(=10\times\left(-201,43\right)\)

\(=-2014,3\)

\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)

Th1:

\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)

\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)

( 1 )

Th2: 

\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)

( 2 )

Từ ( 1 ) và ( 2 ), ta có:

\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)

\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)

Th1:

\(2-x>0\Leftrightarrow x>2\)

\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)

( Loại )

Th2:

\(2-x< 0\Leftrightarrow x< 2\)

\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)

=> \(\dfrac{4}{5}< x< 2\)

`(2/3-0,25+2)-(2-5/2+1/4)-(2,5-1/3)`

`= 2/3 -1/4 +2-2+ 5/2 -1/4 -5/2 +1/3`

`= (2/3 +1/3) +(-1/4 -1/4) + (2-2) + (5/2-5/2)`

`= 3/3 + (-1/2) + 0 + 0`

`= 1 +(-1/2)`

`= 1/2`

\(\left(\dfrac{2}{3}-0,25+2\right)-\left(2-\dfrac{5}{2}+\dfrac{1}{4}\right)-\left(2,5-\dfrac{1}{3}\right)\\ =\dfrac{2}{3}-0,25+2-2+\dfrac{5}{2}-\dfrac{1}{4}-2,5+\dfrac{1}{3}\\ =\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{5}{2}-2,5\right)+\left(2-2\right)+\left(-\dfrac{1}{4}-0,25\right)\\ =\dfrac{3}{3}+\left(2,5-2,5\right)+0+\left(-\dfrac{1}{4}-\dfrac{1}{4}\right)\\ =1+0+0+\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\)

P = (1 + \(\dfrac{1}{1.3}\)).(1 + \(\dfrac{1}{2.4}\)).(1 + \(\dfrac{1}{3.5}\))....(1 + \(\dfrac{1}{2020.2022}\))

P = \(\dfrac{1.3+1}{1.3}\). \(\dfrac{2.4+1}{2.4}\).\(\dfrac{3.5+1}{3.5}\)....\(\dfrac{2020.2022+1}{2020.2022}\)

P=\(\dfrac{\left(2-1\right)\left(2+1\right)+1}{1.3}\).\(\dfrac{\left(3-1\right)\left(3+1\right)+1}{2.4}\)...\(\dfrac{\left(2021+1\right).\left(2022-1\right)+1}{2020.2022}\)

P = \(\dfrac{2.2}{1.3}\).\(\dfrac{3.3}{2.4}\).\(\dfrac{4.4}{3.5}\)....\(\dfrac{2021.2021}{2020.2022}\)

P = \(\dfrac{2.2021}{2022}\)

P = \(\dfrac{2021}{1011}\) 

\(a,\dfrac{1}{2023}>0;-\dfrac{5}{2024}< 0\\ Nên:-\dfrac{5}{2024}< 0< \dfrac{1}{2023}\Rightarrow-\dfrac{5}{2024}< \dfrac{1}{2023}\\ b,\dfrac{678}{876}< 1;\dfrac{987}{789}>1\\ Nên:\dfrac{678}{876}< 1< \dfrac{987}{789}\Rightarrow\dfrac{678}{876}< \dfrac{987}{789}\)

\(c,\dfrac{535353}{585858}=\dfrac{535353:10101}{585858:10101}=\dfrac{53}{58}=1-\dfrac{5}{58}\\ \dfrac{301}{306}=1-\dfrac{5}{306}\\ Vì:\dfrac{5}{58}>\dfrac{5}{306}\Rightarrow1-\dfrac{5}{58}< 1-\dfrac{5}{306}\\ Nên:\dfrac{535353}{585858}< \dfrac{301}{306}\)

\(d,\dfrac{9}{71}=\dfrac{9.3}{71.3}=\dfrac{27}{213}\\ Vì:\dfrac{27}{213}< \dfrac{27}{211}\\ Nên:\dfrac{9}{71}< \dfrac{27}{211}\)

`# \text {04th5}`

\(M=\dfrac{1}{3} - (6 \dfrac{1}{4} - 3) + \dfrac{2}3 \\ = (\dfrac{1}3 + \dfrac{2}3) - (\dfrac{25}4 - 3) \\ = 1 - \dfrac{13}4 \\ = \dfrac{-9}4\)

M = nhé.

Gọi \(\dfrac{9}{x}< \dfrac{9}{y}\left(x>y;x\&y\inℤ\right)\) là 2 phân số cần tìm

Theo đề bài ta có :

\(-\dfrac{11}{13}< \dfrac{9}{x}< \dfrac{9}{y}< -\dfrac{11}{15}\left(x>y\right)\)

\(\Rightarrow-\dfrac{15}{11}< \dfrac{y}{9}< \dfrac{x}{9}< -\dfrac{13}{11}\)

\(\Rightarrow-\dfrac{135}{99}< \dfrac{11x}{99}< \dfrac{11y}{99}< -\dfrac{117}{99}\)

\(\Rightarrow-135< 11x< 11y< -117\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-12;-11\right)\right\}\)

Vậy 2 phân số đó là \(\left(x;y\right)=\left(-\dfrac{9}{12};-\dfrac{9}{11}\right)\)

3x - 12/5 = -0,6

3x = -0,6 + 12/5

3x = -0,6 + 2,4

3x = 1,8

x = 1,8 : 3

x = 0,6

`# \text {04th5.}`

$3x - \dfrac{12}{5} = -0,6$

$\Rightarrow 3x = -0,6 + \dfrac{12}{5}$

$\Rightarrow 3x = \dfrac{9}{5}$

$\Rightarrow x = \dfrac{9}{5} \div 3$

$\Rightarrow x = \dfrac{3}{5}$

Vậy, $x = \dfrac{3}{5}.$