Số tiền Hoa mua gói dâu tây là: \(400000\times\dfrac{1}{3}=\dfrac{400000}{3}\) (đồng)

Số tiền Mai mua nước ngọt là: \(250000\times\dfrac{1}{2}=125000\) (đồng)

Do \(\dfrac{400000}{3}>125000\) nên Mai hết nhiều tiền hơn

1) \(\dfrac{1}{8}x^3-8=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2-x+4\right)\)

2) \(5x^2+10xy+5y^2=5\left(x^2+2xy+y^2\right)=5\left(x+y\right)^2\)

3) \(10x^3-10a=10a\left(a^2-1\right)=10a\left(a-1\right)\left(a+1\right)\)

4) \(2x^3+16=2\left(x^2+8\right)=2\left(x+2\right)\left(x^2-2x+4\right)\)

5) \(x^2-5x+6=\left(x^2-3x\right)-\left(2x-6\right)=x\left(x-3\right)-2\left(x-3\right)=\left(x-2\right)\left(x-3\right)\)

6) \(x^2+5x+6=\left(x^2+3x\right)+\left(2x+6\right)=x\left(x+3\right)+2\left(x+3\right)=\left(x+2\right)\left(x+3\right)\)

7) \(x^2-7x+12=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)

8) \(x^2+x-12=\left(x^2-3x\right)+\left(4x-12\right)=x\left(x-3\right)+4\left(x-3\right)=\left(x+4\right)\left(x-3\right)\)

9) \(x^2+x-20=\left(x^2-4x\right)+\left(5x-20\right)=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)

a) \(x.\left(\dfrac{-1}{2}\right)^5=\left(\dfrac{-1}{2}\right)^7\)

\(\Rightarrow x=\left(\dfrac{-1}{2}\right)^7:\left(\dfrac{-1}{2}\right)^5\\ \Rightarrow x=\left(\dfrac{-1}{2}\right)^2=\dfrac{1}{4}\)

b) \(\left(\dfrac{-3}{4}\right)^8:x=\left(\dfrac{-3}{4}\right)^6\)

\(\Rightarrow x=\left(\dfrac{-3}{4}\right)^8:\left(\dfrac{-3}{4}\right)^6\\ \Rightarrow x=\left(\dfrac{-3}{4}\right)^2=\dfrac{9}{16}\)

Ta có 4x - 4x = 0 em ạ

Do tia Ot nằm trong góc mOn nên ta có \(\widehat{tOm}+\widehat{tOn}=\widehat{mOn}\)

\(\Rightarrow\widehat{tOm}=\widehat{mOn}-\widehat{tOn}=150^0-70^0=80^0\)