\(a,\) \(x.0,\left(2\right)+0,\left(3\right)=0,\left(77\right)\)

⇔ \(x.2.0,\left(1\right)+3.0,\left(1\right)=77.0,\left(01\right)\)

⇔ \(2x.\dfrac{1}{9}+3.\dfrac{1}{9}=77.\dfrac{1}{99}\)

⇔ \(2x.\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{7}{9}\)

⇔ \(2x.\dfrac{1}{9}=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{4}{9}\)

⇔ \(2x=\dfrac{4}{9}:\dfrac{1}{9}=4\)

⇔ \(x=4:2=2\)

Vậy \(x=2\)

\(b,\) \(0,\left(153\right):0,\left(123\right)=1\dfrac{10}{41}.x\)

⇔ \(153.0,\left(001\right):\left[123.0,\left(001\right)\right]=\dfrac{51}{41}.x\)

⇔ \(153.\dfrac{1}{999}:\left(123.\dfrac{1}{999}\right)=\dfrac{51}{41}.x\)

⇔ \(\dfrac{17}{111}:\dfrac{41}{333}=\dfrac{51}{41}.x\)

⇔ \(\dfrac{51}{41}=\dfrac{51}{41}x\)

⇔ \(x=\dfrac{51}{41}:\dfrac{51}{41}=1\)

Vậy \(x=1\)

a)x.0,(2)+0,(3)=0,(77)

x.0,(2)=0,(77)-0,(3)

x.0,(2)=0,47

x=0,47:0,(2)

x=0,77

b) 0,(153):0,(123)=1/10/41.x

1,24390=1/10/41.x

x=1/10/41:1,24390

x=1

x + x : 0,2 = 1,35
x * 1 + x * 5 = 1,35
x * ( 1 + 5 ) = 1,35
x * 6 = 1,35
x = 1,35 : 6
x = 0,225

hok tốt nha ^_^

4,5 k nha

là 7 pn àk

minh cung muon biet

Trả lời :

\(0,1+0,2+0,3+...+0,x=4,5\)

\(0,1\times1+0,1\times2+0,1\times3+...+0,1\times x=4,5\)

\(0,1\times\left(1+2+3+...+x\right)=4,5\)

\(1+2+3+...+x=4,5:0,1\)

\(1+2+3+...+x=45\)

\(\left(x+1\right)\times x:2=45\)

\(\left(x+1\right)\times x=45\times2\)

\(\left(x+1\right)\times x=90\)

Ta có : 90 = 9 x 10

Vậy x = 9

Có: \(\left(x-0,2\right)^{10}\ge0;\left(-y-0,3\right)^{20}\ge0\) với mọi x;y

Mà theo đề bài: (x - 0,2)¹⁰ + (-y - 0,3)²⁰ = 0

\(\Rightarrow\begin{cases}\left(x-0,2\right)^{10}=0\\\left(-y-0,3\right)^{20}=0\end{cases}\)\(\Rightarrow\begin{cases}x-0,2=0\\-y-0,3=0\end{cases}\)\(\Rightarrow\begin{cases}x=0,2\\y=-0,3\end{cases}\)

Vậy x = 0,2; y = -0,3

ta có :(x - 0,2 )¹⁰ lớn hơn hoặc bằng 0 vs mọi x

(-y - 0,3 )²⁰ lớn hơn hoặc bằng 0 với mọi y

=>( x- 0,2 )¹⁰ + (-y - 0,3 )²⁰ =0

<=>\(\begin{cases}\left(x-0,2\right)^{10}\\\left(-y-0,3\right)^{20}=0\end{cases}=0}\Leftrightarrow\begin{cases}x=0,2\\y=-0,3\end{cases}\)