Đặt số đo các góc lần lượt là: a, b, c (độ)

Ta có: a + b + c = 180 độ

Áp dụng tính chất dãy tỉ số = nhau, ta có:

\(\dfrac{a}{5}=\dfrac{b}{3}=\dfrac{c}{2}=\dfrac{a+b+c}{5+3+2}=\dfrac{180}{10}=18\)

\(\Rightarrow\dfrac{a}{5}=18\Rightarrow a=90\)

\(\Rightarrow\dfrac{b}{3}=18\Rightarrow b=54\)

\(\Rightarrow\dfrac{c}{2}=18\Rightarrow c=36\)

  \(\dfrac{x}{4}\) = \(\dfrac{3y}{9}\) ; \(\dfrac{x}{2}\) = \(\dfrac{2z}{10}\) ⇒ \(\dfrac{x}{4}\) = \(\dfrac{2z}{20}\) 

⇒ \(\dfrac{x}{4}\) = \(\dfrac{3y}{9}\) = \(\dfrac{2z}{20}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{4}\) = \(\dfrac{3y}{9}\) = \(\dfrac{2z}{20}\) = \(\dfrac{x+3y-2z}{4+9-20}\) = \(\dfrac{36}{-7}\)

\(x\) = - \(\dfrac{144}{7}\)

y = - \(\dfrac{144}{7}\) : 4 \(\times\) \(\dfrac{9}{3}\) = - \(\dfrac{432}{28}\)

z = - \(\dfrac{144}{7}\) : 2 \(\times\) \(\dfrac{10}{2}\) = - \(\dfrac{720}{14}\)

\(a,\dfrac{2}{3}n+\dfrac{3}{4}n=\dfrac{17}{6}\\ \Leftrightarrow\dfrac{17}{12}n=\dfrac{17}{6}\\ \Leftrightarrow n=2\\ b,\dfrac{1}{5}n+\dfrac{7}{15}=\dfrac{1}{3}n\\ \Leftrightarrow\dfrac{2}{15}n=\dfrac{7}{15}\\ \Leftrightarrow n=\dfrac{7}{2}\\ c,\dfrac{4}{21}-\dfrac{2}{3}n=\dfrac{3}{7}n\\ \Leftrightarrow\dfrac{23}{21}n=\dfrac{4}{21}\\ \Leftrightarrow n=\dfrac{4}{23}\)

Ngày thứ nhất bán được:

\(120\cdot25\%=30\left(kg\right)\)

Số đường còn lại:
\(120-30=90\left(kg\right)\)

Số đường bán trong ngày thứ hai:

\(\dfrac{4}{9}\cdot90=40\left(kg\right)\)

Số đường bán trong ngày thứ ba:

\(120-30-40=50\left(kg\right)\)

50kg

\(1,\left(5\right)=\dfrac{14}{9}\)

14/9 nha

\(25^{2x}:5^x=125^2\)

\(\Rightarrow5^{4x}:5^x=\left(5^3\right)^2\)

\(\Rightarrow5^{4x-x}=5^6\)

\(\Rightarrow5^{3x}=5^6\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

x = 2

`@` `\text {Ans}`

`\downarrow`

`a,`

`5.125.25 \div 5^6`

`=`\(5\cdot5^3\cdot5^2\div5^6\)

`=`\(5^{1+3+2-6}=5^{6-6}=5^0=1\) 

`b,`

\(2^{14}\div\left(2^6\cdot32\right)\)

`=`\(2^{14}\div\left(2^6\cdot2^5\right)\)

`=`\(2^{14}\div2^{11}=2^3\)

`c,`

`3.3^5\div 27`

`=`\(3\cdot3^5\div3^3\)

`=`\(3^{1+5-3}\)

`=`\(3^3\) 

`d,`

\(2^{15}\div\left(2^6\cdot32\right)=2^{15}\div\left(2^6\cdot2^5\right)=2^{15}\div2^{11}=2^4\)

`e,`

\(3^2\cdot27\div81=3^2\cdot3^3\div3^4=3\)

`g,`

\(100\cdot1000\cdot10000-10^9=10^2\cdot10^3\cdot10^4-10^9\)

`=`\(10^9-10^9=0\)

`h,`

\(125^4\div5^9=\left(5^3\right)^4\div5^9=5^{12}\div5^9=5^3\)

a) 1

b) 8

c) 27

d) 16

e) 3

g) 0

h) 125

${\left({25}^2\right)}^x:5^x={125}^2$

${625}^x:5^x={125}^2$

${(625:5)}^x={125}^2$

${125}^x={125}^2$

$x=2$

25^2\(x\) : 5^\(x\)  =125²

5\(^{4x}\) : 5\(^x\) = 5⁶

5^\(3x\)       = 5⁶

3\(x\)      = 6

   \(x\)      = 2

\(6-\left|x+2\right|=2x+3\\ \Rightarrow\left|x+2\right|=6-\left(2x+3\right)\\ \Rightarrow\left|x+2\right|=6-2x-3\\ \Rightarrow\left|x+2\right|=3-2x\)

\(\left|x+2\right|=\left\{{}\begin{matrix}3-2x\Leftrightarrow x\ge-2\\-3+2x\Leftrightarrow x< -2\end{matrix}\right.\)

Với \(x\ge-2\Rightarrow x+2=3-2x\Rightarrow x+2x=3-2\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\left(tm\right)\)

Với \(x< -2\Rightarrow x+2=-3+2x\Rightarrow x-2x=-3-2\Rightarrow-x=-5\Rightarrow x=5\left(ktm\right)\)

Vậy \(x\in\left\{\dfrac{1}{3}\right\}\)