B

b

B

\(180-\left(x-45\right)\div2=120\)

\(\left(x-45\right)\div2=180-120\)

\(\left(x-45\right)\div2=60\)

\(x-45=60\times2\)

\(x-45=120\)

\(x=120+45\)

\(x=165\)

\(\left(2009-x\right)\div29+32=100\)

\(\left(2009-x\right)\div29=100-32\)

\(\left(2009-x\right)\div29=68\)

\(2009-x=68\times29\)

\(2009-x=1972\)

\(x=2009-1972\)

\(x=37\)

\(\left(x-2\right)\times3-270\div45=120\)

\(\left(x-2\right)\times3-6=120\)

\(\left(x-2\right)\times3=120+6\)

\(\left(x-2\right)\times3 =126\)

\(x-2=126\div3\)

\(x-2=42\)

\(x=42+2\)

\(x=44\)

\(\left(x+0,2\right)+\left(x+0,7\right)+\left(x+1,2\right)+...+\left(x+4,2\right)+\left(x+4,7\right)=65,5\)

\(\left(x+x+x+..+x+x\right)+\left(0,2+0,7+1,2+...+4,2+4,7\right)=65,5\)

\(10x+\frac{\left(0,2+4,7\right)\times10}{2}=65,5\)

\(10x+24,5=65,5\)

\(10x=65,5-24,5\)

\(10x=41\)

\(\Rightarrow x=4,1\)

`x-0,7<8`

`=>x<8,7`

Mà `8<x`

`=>8<x<8,7`

Nếu `x in ZZ=>x in` $\varnothing$

`x in R=>8<x<8,7`

Ta có: x-0.7<8<x

\(\Leftrightarrow x< 8.7\)

mà 8<x

nên 8<x<8,7

Vậy: S={x|8<x<8,7}

Từ 0,2 đến 4,7 có số số hạng là: \(\left(4,7-0,2\right):0,5+1=10\left(số\right)\)

⇒ Tổng 10 số đó là: \(\dfrac{\left(4,7+0,2\right)\cdot10}{2}=24,5\)

\(\left(x+0,2\right)+\left(x+0,7\right)+...+\left(x+4,7\right)=65,5\\ \Leftrightarrow10x+24,5=65,5\\ \Leftrightarrow10x=41\\ \Leftrightarrow x=4,1\)

Vậy x = 4,1

\(x\%:3+45\%=0,7\)

\(\Rightarrow x\%:3+0,45=0,7\)

\(\Rightarrow x\%:3=0,7-0,45\)

\(\Rightarrow x\%:3=0,25\)

\(\Rightarrow x\%=0,25.3\)

\(\Rightarrow x\%=0,75\)

\(\Rightarrow x\%=75\%\)

\(\Rightarrow x=75\)

Vậy \(x=75\)

\(x\%:3+45\%=0,7\)

\(\Rightarrow\frac{x}{300}+0,45=0,7\)

\(\Rightarrow x:300=0,25\)

\(\Rightarrow x=75\)

đáp án 

x = 75 

hok tốt

=>|0,7-x|=-5,3+11 = 5,7

=> 0,7-x=5,7 hoặc 0,7-x=-5,7

=> x=-5 hoặc x=6,4

k mk nha

|0,7-x|= -5,3 + 11 = 5,6

\(\Rightarrow\orbr{\begin{cases}0,7-x=-5,6\\0,7-x=5,6\end{cases}\Rightarrow\orbr{\begin{cases}x=6,3\\x=4,9\end{cases}}}\)

=>   x + 0,7 = -4    =>       x =   -4,7

bn nguyễn  hà lan anh đã học về phần nguyên chưa vậy

rùi thì pải

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)

\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)

\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)

\(=5-\dfrac{4}{2}\)

\(=5-2\)

\(=3\)

b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)

\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)

\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)

\(=0+0+0+2022\)

\(=2022\)

2) \(0,7^2\cdot x=0,49^2\)

\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)

\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)

\(\Rightarrow x=\left(0,7\right)^2\)

\(\Rightarrow x=0,49\)

b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)

\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)

\(\Rightarrow x=\left(-0,5\right)^5\)

\(\Rightarrow x=-\dfrac{1}{32}\)

2:

a: =>x*0,49=0,49^2

=>x=0,49

b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5

( x + 0,7 )³ = -27

( x + 0,7 )³ = (-3)³ 

=> x + 0,7 = -3

x = -3 -0,7

x = -3,7

Vậy x = -3,7

=))

\(\left(x+0,7\right)^3=-27\)

\(\Rightarrow\left(x+0,7\right)^3=\left(-3\right)^3\)

\(x+0,7=\left(-3\right)\)

       \(x=\left(-3\right)-0,7\)

       \(x=-3,7\)

Vậy \(x=-3,7\)

Chúc bạn học tốt !!!

(x+0,7)^3=3^3

x+0,7=3

x= 3-0,7

x=2,3