\(\left(x-\dfrac{2}{7}\right)^5=32\\ =>\left(x-\dfrac{2}{7}\right)^5=2^5\\ =>x-\dfrac{2}{7}=2\\ =>x=\dfrac{2}{7}+2\\ =>x=\dfrac{16}{7}\)

VI

1 exciting

2 wonderful

3 interested

4 dangerous

5 safety

6 unhealthy

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1a

2c

3b

4a

5d

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1c

2b

3a

4d

5a

#\(yGLinh\)

\(\left(x+\dfrac{4}{3}\right)^4=\dfrac{1}{81}\\ =>\left(x+\dfrac{4}{3}\right)^4=\left(\dfrac{1}{3}\right)^4\)

TH1: `x+4/3=1/3`

`=>x=1/3-4/3`

`=>x=-3/3`

`=>x=-1`

TH2: `x+4/3=-1/3`

`=>x=-1/3-4/3`

`=>x=-5/3`

Vậy: ...

\(\overline{abc4}-\overline{abc}-1111\\ \overline{abc}\cdot10+4-\overline{abc}=1111\\ \overline{abc}\cdot\left(10-1\right)=1111-4\\ \overline{abc}\cdot9=1107\\ \overline{abc}=1107:9\\ \overline{abc}=123\) 

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\(\overline{ab5}=\overline{ab}+230\\ \overline{ab}\cdot10+5=\overline{ab}+230\\ \overline{ab}\cdot10-\overline{ab}=230-5\\ =>\overline{ab}\cdot\left(10-1\right)=225\\ =>\overline{ab}\cdot9=225\\ =>\overline{ab}=225:9\\ =>\overline{ab}=25\)

                           Giải:

Diện tích của hình vuông ABCD là: 12 x 12 : 2 = 72 (m²)

Diện tích của hình tròn tâm O bán kính OK là:

  KG.EH.3,14 : 4 = \(\dfrac{1}{4}\)KG.EH = \(\dfrac{3,14}{4}\)S_ABCD = 72 x \(\dfrac{3,14}{4}\) = 56,52 (m²)

     ⇒ MP.QN.3,14:4 = 56,52

    ⇒ MP.QN = 56,52 x 4 : 3,14 = 72

 Diện tích hình vuông MNPQ là:

   MP.QN : 2 = 72: 2 = 36 (m²)

Diện tích phần gạch chéo là: 

     56,52  - 36  = 20,52 (m²)

    Số tiền ông Nam phải chi trả cho nghệ nhân trang trí phần diện tích gạch chéo là:

            250 000 x 20,52  = 5 130 000 (đồng)

Kết luận:...

\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

2. She lives in Ho Chi Minh City

4. No she doesn't 

 Cho mình xin 1 tick với ạa

\(B=4\cdot2^7:\left(3^{11}\cdot\dfrac{1}{9}\right)\\ =4\cdot2^7:\left(3^{11}\cdot\dfrac{1}{3^2}\right)\\ =4\cdot2^7:\dfrac{3^{11}}{3^2}\\ =4\cdot2^7:3^9\\ =\dfrac{2^2\cdot2^7}{3^9}\\ =\dfrac{2^9}{3^9}\\ =\left(\dfrac{2}{3}\right)^9\)

\(b)\left\{{}\begin{matrix}x-\dfrac{y}{2}=\dfrac{1}{2}\\\dfrac{x}{3}-2y=-\dfrac{5}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-6y=-5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\2x-12y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}11y=11\\2x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\2x=1+1=2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)

\(c)\left\{{}\begin{matrix}5x-0,7y=1\\-10x+1,4y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\-5x+0,7y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-0,7y=1\\5x-0,7y=1\end{matrix}\right.\) 

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