1) stayed

2) went

3) had

4) was

5) visited

6) were

7) bought

8) saw

9) ate

10) talked

11) returned

12) thought/ was

13) worked

14) studied

15) travelled

16) watched

17) bought

18) thought/ was

19) cooked

20) wrote

1) weren't / were

2) was/ wasn't

3) Was

4) were

5) wasn't

6) was

7) Was/ wasn't

8) were/ were

9) Were/ weren't

10) Was/ was

\(2x^2-6x+8=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}+8=2\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{2}\)

Vì \(2\left(x-\dfrac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\dfrac{3}{2}\right)^2+\dfrac{7}{2}\ge\dfrac{7}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{2}\)

Vậy GTNN của \(2x^2-6x+8\) là \(\dfrac{7}{2}\) khi và chỉ khi \(x=\dfrac{3}{2}\)

a) \(5x^2y-20xy+20y=5y\left(x^2-4x+4\right)=5y\left(x-2\right)^2\)

b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)

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

d) \(7x^3-28x^2+28x=7x\left(x^2-4x+4\right)=7x\left(x-2\right)^2\)

a) \(\left(2x^4-3x^3-3x^2-2+6x\right):\left(x^2-2\right)=2\left(x^2-\dfrac{3}{2}x+\dfrac{1}{2}\right)\left(x^2-2\right):\left(x^2-2\right)=2x^2-3x+1\)

\(2^m+2^n=2^{m+n}\Rightarrow2^m+2^n=2^m.2^n\Rightarrow2^m.2^n-2^m-2^n=0\Rightarrow2^m\left(2^n-1\right)-\left(2^n-1\right)-1=0\Rightarrow\left(2^n-1\right)\left(2^m-1\right)=1=1.1\)Vì m,n là số tự nhiên nên:\(2^m-1=2^n-1=1\Rightarrow2^m=2^n=2\Rightarrow m=n=1\)

a) \(n^3-n=n\left(n^2-1\right)=\left(n-1\right)n\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên chia hết cho 3

b) \(n\left(n-1\right)\left(2n-1\right)=n\left(n-1\right)\left(n+1+n-2\right)=\left(n-1\right)n\left(n+1\right)+\left(n-2\right)\left(n-1\right)n\)Ta có: \(\left(n-1\right)n\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên có một số chia hết cho 2 và một số chia hết cho 3, mà(2,3)=1 nên \(\left(n-1\right)n\left(n+1\right)⋮6\) 

Tương tự ta cũng được \(\left(n-2\right)\left(n-1\right)n⋮6\)

\(\Rightarrow\left(n-1\right)n\left(n+1\right)+\left(n-2\right)\left(n-1\right)n⋮6\)

\(\Rightarrow n\left(n-1\right)\left(2n-1\right)⋮6\left(đpcm\right)\)

1)\(32^{-n}.16^n=\dfrac{1}{1024}\Leftrightarrow2^{-5n}.2^{4n}=\dfrac{1}{2^{10}}\Leftrightarrow2^{-n}=\dfrac{1}{2^{10}}\Leftrightarrow2^n=2^{10}\Leftrightarrow n=10\)

a) \(120.15+12.470+12.38.10=120\left(15+47+38\right)=120.100=12000\)b) \(13.58.4+32.26.2+52.10=52\left(58+32+10\right)=52.100=5200\)c) \(3,2.990+32=3,2\left(990+10\right)=3,2.1000=3200\)d) \(45\left(13+78\right)+9\left(87+22\right).5=45.91+45.109=45\left(91+109\right)=45.200=9000\)e) \(\left(24+72+36+60\right):12=\left(24+72\right).\left(36+60\right):12=96.2:12=192:12=16\)f) \(\left(2+4+6+...+100\right)\left(36.333-108.111\right)=\left(2+4+6+...+100\right)\left(36.3.111-108.111\right)=\left(2+4+6+...+100\right)\left(108.111-108.111\right)=\left(2+4+6+...+100\right).0=0\)